A man standing at the top of a 20 meter high building throws an apple straight upward with a velocity of 9 m/s. The apple falls subsequently falls to the ground.

Question

A man standing at the top of a 20 meter high building throws an apple straight upward with a velocity of 9 m/s. The apple falls subsequently falls to the ground. What is the apple's speed right before it reaches the ground?

Answer 1

This is a 2-part question: In order to know the final velocity, we first need to find the total height the apple fell from. That will be 20 meters PLUS whatever height was added by the upwards throw.

To calculate that upwards throw height, let's start by looking at our variables. We know that the upwards initial velocity (v₀) is 9 m/s, the peak final velocity (v_f) will be 0 m/s (where it stops going up and starts coming down), and gravity (a) equals -9.8 m/s/s. We don't know the distance (d) travelled to get from 9 to 0 m/s, and we also don't know how much time (t) it took.

Next, let's find a kinematics formula that uses the variables we're given and only one unknown that we can solve for. Kinematic Equations and Free Fall (physicsclassroom.com) or What are the kinematic formulas? (article) | Khan Academy. I see that v_f² = v₀² + 2(a)(d), which is perfect. When we substitute our givens, we get that 0 = 9² + 2(-9.8)(d). Therefore, d = -9² / (2 * -9.8) = -81 / -19.6 = 4.13 meters.

So, now we know that the total distance fallen (d) is 20 + 4.13 = 24.13 meters. We know that gravity (a) is still -9.8m/s/s, and v₀ = 0 when it first starts to come back down. We need to solve for v_f, and that same formula can help us here v_f² = v₀² + 2(a)(d). Substituting our givens we get that v_f² = 0² + 2(-9.8)(24.13) = 472.95, therefore v_f = √472.95 = 21.75m/s.