talk about a projectile which is thrown from the ground so here we have a projectile and this projectile is launched at an angle theta which is given that angle is 30 meter per second squared 30 pokeman 30 degree angle and this velocity has a velocity at which the projectile is launched is 25 meter per second thus the only the two information given and now what we need to find it out is how long the the particle takes to reach to the maximum height so we need to find out the time it takes to reach to the maximum height and then what is the horizontal velocity and the vertical velocity at this at the peak and how long does it take the particle to reach to the ground and again what are the horizontal and the vertical velocity at the ground but what angle it hits the ground what is the horizontal range and what is the position of the particle at time equals 2.5 seconds so this one problem has multiple sub problems into it and I tell you I must tell you but if you solve this problem after understanding or if you can solve this problem by yourself that means you understand the projectile motions completely okay so how are you going to do it and you do not need any formula here you do not need to memorize the only thing you need to memorize or keep in mind is the basic kinematic equations you do not need to memorize the projectile the formula okay so let's do it so this one is the V nod the X component will be V naught cosine theta because this angle is Theta and the Y component will be V naught or V zero sine theta that's what I have written here this is the initial horizontal velocity which is the V zero X and if we plug in the value of V 0 age 25 cosine theta of course I'm 30-degree if you solve it it'll be twenty one point six five meter per second and now the vertical velocity initial vertical velocity or the velocity along the Y direction will be V zero sine theta his 0 is 25 and sine theta is sine 30 degree so you solve it he'll get 12 point 5 meter per second so we now figure it out the horizontal velocity and they vertical velocity okay now what I'm going to know now there is no acceleration along the x axis why is that there's only one force acting on to the system or acting onto the projectile at any point at any point the force is only acting along the Y direction and that force is the force due to the gravity which is mg that's the only force acting there's no any force there's no force on the particle along the x direction if there is no force acting along acting on the particle along x direction that means the particle of velocity will not change the velocity will change if there is a force acting on to it and I and as there is no force acting along the x component the velocity at all point will remain exactly the same that means the acceleration would be 0 so we have acceleration along the x axis 0 and along Y axis it is acceleration due to the gravity ok now we are going to calculate time to reach the peak and the maximum height the maximum height is this value from the ground to the peak that's the maximum height so we're going to write down the equation along the vertical motion or along the Y direction and here we have you our children this equation to calculate that time equals to u minus GT in wide negative sign here because as the particle is going up the velocity and the extra and acceleration are in opposite direction that's the reason we choose D the negative Direction negative sign here or in other words simply understand keep in mind that if the particle is going up you just write down the negative if the particle is going down then the velocity would increase and he had to write down the positive value the positive G and the pic had to pick that velocity zero so whenever you throw a particle into the air it will reach to a height and the velocity starts decreasing and at the peak or at the top that particle velocity reaches to zero that's all that's the reason it is at home and they initially did you a stands for the initial velocity which is V zero Y one is e and the T and the T is the time it takes to reach to the pick so if they solve for the time or the get is V zero Y over G and what is the initial vertical velocity 12.5 what is the value of G which is 9.8 so the time is now one point two eight second that will set it at this time it takes to reach to the pick now again we taking the vertical direction or motion along the y axis I'm going to solve for the maximum height this time we're going to use this equation via squirt it costs to us squared minus 2 gee H okay and you know now what is the V if he is the final velocity and the final velocity at the pig is 0 and U is the M so velocity and what is the initial velocity we already have calculated which is V zero Y and here now 2 G H and H is the maximum height and if we solve for the maximum height we get officially square over 2 G and here we already have calculated each ly which is twelve point five so twelve point five square to and G value is nine point eight and if you solve it what I get is seven point nine seven meter so the maximum height this height is seven point nine seven meter okay now the third thing is what are the velocities at the pit this is now redundant questions we already have done it but okay let's do it one more time the velocities at the peak so you see now the horizontal velocity does not change at any position it remembers exactly the same horizontal velocity so even at the peak the horizontal velocity or the velocity along the x axis is exactly equal to the initial horizontal velocity which we already have calculated to be twenty one point six five okay and at the peak the vertical velocity is zero okay the particle is slowing them slowing down and at the pic this velocity zero the vertical velocity but it still it has the horizontal velocity okay so we now got this answer now time to reach to the ground how do we calculate time to risk to the ground we have calculated time to risk to the peak from ground to the peak and if we ignore air resistance it takes exactly the same time from peak to the ground so the total time this total time would be simply equal to double at the time it takes to reach the ground this time and this time is exactly the same okay we already have calculated at this time here so the total time would be two of the time it has to reach to the feet and the time to reach to the peak is one point two eight seconds so time to reach to the ground is two times time to reach to the pig and the time is now two point five six five six seconds just do the math now the velocities at the ground what are the velocities at the ground you know now I am so sorry you know the answer now so the horizontal velocity does not change so you know you you know the velocity the horizontal velocity which is exactly equal to the initial velocity and the vertical velocity would be exactly this velocity but the only difference is that you recently changed now this one was going up and this is going down so it's exactly the same thing so Monica tear only the toxin has not changed so the so we now know the answer for this one to the vertical velocity at the ground is exactly twenty to one point six five which is the initial velocity and the velocity along the ground it's exactly the same velocity okay and the reason is if we did all the resistance the gravitational force is a conservative force so if you throw a particle into the air and be just holding in your hand let's say you throw there say you are here and this is your hand and you throw a particular into the air with twenty meter per second on brittonic path when it comes to your hand it will come to your hand exactly a 20 meter per second okay so at the ground the floor should be exactly the same thing just by symmetry as well you already know what the velocity the vertical velocity at this point and that this point would be exactly the same at this point and at this point would be exactly the same just by symmetry okay so even at this point at this point would it be exactly the same I remember that the victim would of course be changed opposite now the last thing we need to calculate is not the last thing I think the angle what is the angle at which it hits the ground well I get her it would hit the ground exactly at the same angle at which it was launched but you can still calculate it how do you going to calculate the tangent theta is so if this is a horizontal velocity and this is the vertical velocity this is the VG X and this is the VG y then this angle theta would be calculated by tangent theta is VG Y the Y component divided by the X component that's how you calculate the angle and the VG y is 12.5 and if AZ X already had calculate twenty one point six five they were solved for theta you got the 30 degree you don't have to solve it just by symmetry again the angle at which it hits the ground put a exactly this angle theta okay now we need to find out the horizontal range so hota how to calculate the horizontal relationship the horizontal range I'm just taking the motion now along the extraction s so we're going to use this equation to calculate the horizontal range S is a horizontal range that horizontal range in this picture is from ground to ground this distance is the horizontal range here okay and this formula U is the initial velocity so I'm just writing down the V zero X and the time this time is the total time it remains in the air so I have written down two times T the T here is time it takes to reach to the pig so the total time in tournaments in the air is two times T and remember acceleration along the x axis is zero so there is no acceleration along the x axis okay so life is now lot simpler now we know the value v GX twenty one point six five two and the time the time to reach to the pig we already have calculated which is one point or you can simply juice two point five six so the time to reach to the pig is one point two eight so once I plug in called the value the horizontal drains that I get is fifty five point four two meter now the last thing is what is the position of the particle at time because 2.5 second so in order to calculate the x position it's pretty simple there is no acceleration along the x axis so simply you use x equals to V X V zero the velocity turns of time and the velocity along x-axis is twenty one point six five and the time is 0.5 seconds so the x position is ten point eight three meter and now we're going to calculate the Y position but now we have to use this equation here because there is an acceleration as well and the negative sign because the particle is thrown up so this is the Y position because s stands for the height and U is the initial vertical velocity time half GT square so V zero Y the initial vertical velocities two pond5 the time is 0.5 1/2 g is 9.8 and the time is point 2 5 squared everything is constant that's a number here so when I solve it what I get is y is point five point zero two five meter so the height is this height is five point zero two five meter so this is the position so in this way you can calculate the position at any time okay so the key for solving this kind of problem is to solve in a tooth independent coordinate system one along x-axis and one along the y axis okay so you can also use the formula for the horizontal range but I won't encourage you to do that just write down the basic kinematic equations and solve from there and at the last what I would recommend you to solve exactly the same problem but at a different angle let's make it sixty degree I'll solve exactly the same problem and write down your answer in the comment section below and at the hand do not forget to Like share and subscribe the channel thank you very much