Question

3.50 q: a ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. if the ball hits the ground 4.0 seconds later, approximately how high is the cliff? (a) 6.0 m (b) 39 m (c) 78 m (d) 96 m

Explanation:

Step1: Identify vertical - motion parameters

The ball is thrown horizontally, so the initial vertical velocity $u_y = 0\ m/s$. The acceleration due to gravity $g= 9.8\ m/s^{2}\approx10\ m/s^{2}$, and the time of flight $t = 4.0\ s$.

Step2: Use the vertical - displacement formula

The formula for vertical displacement $h=u_y t+\frac{1}{2}gt^{2}$. Since $u_y = 0\ m/s$, the formula simplifies to $h=\frac{1}{2}gt^{2}$.
Substitute $g = 10\ m/s^{2}$ and $t = 4\ s$ into the formula: $h=\frac{1}{2}\times10\times4^{2}$.
$h = 5\times16=80\ m$. The closest value to $80\ m$ among the options is $78\ m$.

Answer:

C. 78 m