in this problem I will talk about a projectile which is drawn from height but at an angle-theta so and what you need to find out the same thing is in the previous problem you need to find out the horizontal rays how long does it remain in the air what are the horizontal velocity and the vertical velocity at what angle it hits the ground so all those information with this information with the initial speed of 10 meter per second and the angle theta and the height will just calculate everything so here are the given information the initial velocity is given which is 10 meter per second the angle theta is given which is 45 degree and the height is 2 meter so once we know the initial velocity of the angle theta we should be able to calculate the the X component of the velocity the X component of the velocity is a v-0 cosine theta because this angle is Theta and the y component would be equal to v-0 sine theta so now let's just find the number V 0 X is V zero cosine theta the initial velocities 10 cosine theta is 45 degree so what you get is 7.07 meter per second does the initial horizontal velocity the same way the initial vertical velocity would reveal 0 sine theta and if you plug in all the numbers you'll get 7.07 which makes sense because theta is equal to 45 degree so X component and Y component has to remain exactly the same okay so now let's find out the time of flight what does that mean that means when you launch the projectile how long does it remain in the air or the time this you need to find out this total time the time from this height from this position till it hits the ground I'm going to find it out so for in order to from the time we're just going to take the motion along the x-axis forget about everything just thinks you have a ball here and the ball is launched with this velocity v-0 sine theta what will happen this ball will reach to the pick and then it come back to the ground so all the motion you have to think about is wounded along the y-axis okay and all the initial information about the hello the y-axis is provided okay so we're going to use this formula here the kinematic equation they say the same as this equation here Y is the final height Y naught is your initial height V 0 is the initial velocity t is the time half GT square so the final height Y so this is here I'm just using this as my coordinate system so when the ball it hits the ground my final height will be the zero right and my initial height is a two meter so the final height is zero and his silhouette is 2 meter and the initial velocity we already have calculated 7.07 times t minus half G is 9.8 T Square and I'm just going to rearrange this equation and again the negative sign here because these because the particle is moving upward so if I'm rearrange this equation nine point eight divided by two will be four point ninety square then 7.07 and minus two because I'm just taking this to the opposite side okay so rearranging you'd get this equation this is exact this is a quadratic equation and you know how to solve a quadratic equation I have written down the formula here how to solve for a quadratic equation so so you can say well if you do this equation to solve this quadratic equation the time you'll get is to time because this is a quadratic equation you'll get two time values the one is one point six nine second and the other is 0.24 second what are the negative time means does it make any sense well or the first level it doesn't make any sense the negative time the time cannot be negative but you need to understand the math behind it and the number does not come by itself there is something hidden behind it if you are in question you should be able to find it out what are the negative time means I'll explain it here although have written down the negative time does not make any physical sense in this particular problem but in fact it does so what is the negative time how do you get this negative time point to four seconds let's come back here so the negative time you will get this point two four second if the state of throwing this value at a at this angle if you throw this projectile exactly at the same angle in this direction or if your vertical velocity is coming downward then you'd get this value point two four second that's the negative valid means but in this particular problem we are not interested we not have not lost a projectile in the downward direction but rather upward Rison at an angle so this time here doesn't make sense is not feasible so only time feasible is one point six nine second but you now got what is the meaning of this one okay and titta launching now if the projectile was launched at this angle then the time you will get would be this number point two four second that's the meaning the secret meaning here okay so now let's calculate the horizontal distance a horizontal distance I have told you multiple times but there is no acceleration along the x-axis okay so this equation x equals to V naught X plus half ay T Square a is the acceleration along x-axis what is the acceleration along x-axis general well that there is no acceleration along the x axis because there is no force acting along the x axis the only force acting along the x axis active on the projectile is the force due to the gravity which is only along the X to the Y direction there's no force along the x axis as there is no force acting along the x axis there's no acceleration along the x axis as well okay so now coming back to this equation X is the horizontal range V zero X is the initial horizontal velocity which we have calculated to be seven point zero seven okay and the time is one point six nine this is a time in two dimension the air which would have gotten from this equation and this is zero because the acceleration is zero so the horizontal breeze we have calculated is eleven point nine two five meter and now let's calculate the what are the velocity when it hits the ground the horizontal velocity and vertical velocity needless to say that the horizontal velocity will remain exactly the same and what's the answer the answer is there's no force among the x-axis so the velocity was remain exactly the same along the x direction so at the ground the horizontal velocity would be exactly the same now let's calculate what is the velocity along the y axis when it hits the ground so I'm going to use this equation Z equals to u minus GT there again the negative sign here because we have long as the projectile and the upward direction this is the final velocity so the final velocity was V Gy and the u.s. stands for the initial velocity and it's still vertical velocity the V 0 y 0 stands for the initial G times time so we are interested in finding the V Z 1 the initial vertical velocity is again given or we have calculated which is seven point zero seven and now G and the time we are will just calculate the time as well this time 1.69 second a turn it takes to reach to the ground we got a negative number what is the negative number mean again the negative number means the velocity is now coming in the negative y-direction which makes perfect sense because when we launch the projector at this angle it agrees to the crowd and now the direction is flipped the velocity would be coming in the town water except that's what the negative sign means okay so the negative sign shows that the velocity direction is vertically downward now let's find out the angle so how do you calculate the angle again so okay let's come back here one more time and do not get confused at this point this theta and this would be different if you calculate theta on the same lever it would be exactly this angle and this angle it'll be exactly the same on a sale level ground but now these two are not at the same level so this angle theta and this angle alpha would be different okay then now you you can also get a sort of idea too because the vertical horizontal velocity is remaining exactly the same and the vertical velocity is increasing okay so there is more and more component so the Alpha will increase so okay so how do you find out this angle alpha here so if you look at this time this is the X component and this is the Y component so the Alpha and this is the net velocity if you look at this triangle in this triangle the tangent of alpha would be Y component V is your Y or V DX that's how you find out the angle let's use this one the tangent of alpha is VG y over VG X and VG y is negative we already have calculator 9 point 4 9 and VG X is seven point zero seven and alpha is negative fifty three point three degree two things you have to understand here the one is alpha is negative and does it make sense yes it does make sense because now the velocity is flipped so it is not moving forward it is going downward that's what it means and the angle is greater than 45 degree which makes perfect sense because the vertical velocity is increasing while the horizontal velocity is it amending the same so the particle will be mourning will be getting more vertical and vertical so it has to be greater than 45 degree and now the speed at which it hits the ground is or the total magnitude of the velocity is given by this haruna here just by the Pythagoras theorem the VG x squared and PG y squared we have calculated the vzx 7.07 and the VG y is nine point four nine if you solve it you'll get eleven point eight three meter per second so this is the speed this is an angle okay so this is it what I would like to recommend at the end of this lecture is salt is exactly solve this problem but now not at forty-five degree but at 30 degree and tell me what the answer you get write down your answer in the comment section okay and at the end again I encourage you to subscribe the channel so that you'll get all the videos directly to you okay thank you so much