Question

current attempt in progress a ball is dropped from rest from the top of a building and strikes the ground with a speed ( v_f ). from ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. the initial speed of the second ball is ( v_0 = v_f ), the same speed with which the first ball eventually strikes the ground. ignoring air resistance, decide whether the balls cross paths (a) at half the height of the building, (b) above the halfway point, or (c) below the halfway point. (\bigcirc) (a) (\boldsymbol{\bigcirc}) (b) (\bigcirc) (c)

Explanation:

1. Let the height of the building be \( h \) and acceleration due to gravity be \( g \). For the first ball (dropped from rest), using \( v_f^2 = 2gh \) (from kinematic equation \( v^2 = u^2 + 2as \), where \( u = 0 \), \( a = g \), \( s = h \)).
2. For the first ball's position at time \( t \): \( y_1 = h - \frac{1}{2}gt^2 \) (downward is positive, starting from height \( h \)).
3. For the second ball (thrown upward with \( v_0 = v_f \)): \( y_2 = v_0t - \frac{1}{2}gt^2 = v_f t - \frac{1}{2}gt^2 \).
4. Set \( y_1 = y_2 \): \( h - \frac{1}{2}gt^2 = v_f t - \frac{1}{2}gt^2 \), so \( h = v_f t \). From \( v_f^2 = 2gh \), \( v_f=\sqrt{2gh} \), substituting gives \( t=\frac{h}{\sqrt{2gh}}=\sqrt{\frac{h}{2g}} \).
5. Substitute \( t \) into \( y_1 \): \( y_1 = h - \frac{1}{2}g(\frac{h}{2g}) = h - \frac{h}{4}=\frac{3h}{4} \), which is above \( \frac{h}{2} \).

Answer:

(b) above the halfway point