in this video we'll solve a problem about vertical motion in the presence of gravity Let's first read the problem statement Let's visualize what's happening in this problem We have two stones being thrown upward from the ground but at different times and with different initial velocities The first stone is thrown with an initial velocity of 48.3 ft per second Then 1 second later the second stone is thrown with a higher initial velocity of 96.6 ft pers We want to find out how high they'll be when they reach the same height To solve this problem we need to use the equations of motion for objects under constant acceleration When an object is thrown upward near its surface its height h as a function of time t follows the equation H= v * tus gt ^2 where v kn is the initial velocity and g is the acceleration due to gravity which is 32.2 ft/s squared Now let's set up the equations for our two stones For the first stone the height is h1 = 48.3 * t minus 16.1 * t ^2 Note that I simplified g over 2 to 16.1 For the second stone we need to account for the fact that it's thrown 1 second after the first stone When the first stone has been in the air for time t the second stone has only been in the air for t minus1 seconds So our equation becomes H2 = 96.6 * open parenthesis t minus 1 close parenthesis -6 1 * open parenthesis t minus 1 parenthesis squared Remember that this equation is only valid for t greater than or equal to 1 Since the second stone isn't thrown until 1 second has passed To find when the stones are at the same height we need to set the two height equations equal to each other and solve for t So we have h1 of t equals h2 of t Substituting our expressions we get 48.3t -6.1 t ^2 = 96.6 * open parenthesis t minus 1 close parenthesis -6.1 * open parenthesis t minus 1 parenthesis squared Now let's expand and simplify the right side Let's expand the right side step by step First 96.6 6 * open parenthesis t minus 1 closed parenthesis = 96.6 t - 96.6 Next for the second term we use the formula for open parenthesis t minus 1 close parenthesis 2 which is t ^ 2 - 2t + 1 So we get -16.1 * open parenthesis t^ 2 - 2t + 1 parenthesis Distributing the -6.1 we get -16.1 t ^2 + 32.2t -6.1 So the right side becomes 96.6t - 96.6 - 16.1 t ^2 + 32.2t -6.1 which simplifies to -16.1 t ^2 + 128.8 tus 112.7 Now let's put this back into our original equation We have 48.3 t -6.1 t ^2 = -16.1 t ^2 + 128.8 tus 112.7 Let's move all terms to one side We get 48.3 tus 16.1 t ^2 + 16 1 t ^2 minus 128.8 8 t + 112.7 = 0 Simplifying the t^2 terms cancel out and we're left with 48.3t minus 128.8 t + 112.7 = 0 Combining like terms we get80.5t + 112.7 = 0 Now solving for t we get t = 112.7 / 80.5 which is approximately 1.4 seconds Since this is greater than 1 our solution is valid It occurs after the second stone is thrown Now that we know when they're at the same height let's find what that height is We'll substitute t= 1.4 into either height equation Let's use the equation for the first stone h= 48.3 * 1.4 4 - 16.1 * 1.4^ 2 First we calculate 48.3 * 1.4 which is 67.62 Then 16.1 * 1.4^ 2ar which is 16.1 * 1.96 giving us 31.556 Subtracting we get 67.62us 31.556 which equals approximately 36.06 ft Let's visualize what happens The first stone is thrown at time zero with a velocity of 48.3 ft pers It starts rising but its velocity decreases due to gravity At time equals 1 second the second stone is thrown with a higher velocity of 96.6 ft per second Both stones continue rising but the second stone moves faster At t= 1.4 In 4 seconds they meet at a height of 36.1 ft above the ground After that the second stone continues to overtake the first one Let's summarize our solution We set up the height equations for both stones using the physics of projectile motion We set these equations equal to find when the stones are at the same height After solving we found that the stones meet at t= 1.4 seconds after the first stone is thrown At this time both stones are approximately 36.1 ft above the ground