Questions: Problem 3 Thelma stands on top of a building whose height is H and throws a ball upward. The speed of the ball when it leaves her hand is vT, where the subscript stands for both Thelma and top. Bob is standing at the bottom of the building and at the same instant throws another ball upwards with speed vB. What is the slowest that vB can be if Bob's ball hits Thelma's when Thelma's ball is still on the way up? Express your answer in terms of gravitational acceleration g, H, and vT. (I would recommend that you draw a picture.)

Solution Steps

Thelma throws a ball upward from the top of a building of height \( H \) with an initial speed \( v_T \). Bob throws another ball upward from the ground with an initial speed \( v_B \). We need to find the minimum speed \( v_B \) such that Bob's ball meets Thelma's ball while Thelma's ball is still ascending.

Thelma's ball is subject to gravitational acceleration \( g \) and its motion can be described by the equation: \[ y_T(t) = H + v_T t - \frac{1}{2} g t^2 \] where \( y_T(t) \) is the height of Thelma's ball at time \( t \).

Bob's ball is also subject to gravitational acceleration \( g \) and its motion can be described by: \[ y_B(t) = v_B t - \frac{1}{2} g t^2 \] where \( y_B(t) \) is the height of Bob's ball at time \( t \).

For the balls to collide, their heights must be equal at some time \( t \): \[ H + v_T t - \frac{1}{2} g t^2 = v_B t - \frac{1}{2} g t^2 \] Simplifying, we get: \[ H + v_T t = v_B t \] \[ H = (v_B - v_T) t \] \[ t = \frac{H}{v_B - v_T} \]

Thelma's ball is still ascending if its velocity is positive: \[ v_T - g t > 0 \] Substitute \( t = \frac{H}{v_B - v_T} \): \[ v_T - g \left(\frac{H}{v_B - v_T}\right) > 0 \] \[ v_T (v_B - v_T) > gH \] \[ v_B v_T - v_T^2 > gH \] \[ v_B > \frac{gH + v_T^2}{v_T} \]

Final Answer