Question

A mis-hit golf ball flies straight up in the air. Exactly 4.0 seconds later it lands right next to the tee. How high up did the golf ball go?

Answer 1

Answer:

19.6 m

Explanation:

The total motion of the golf ball lasts 4.0 seconds: since the motion is symmetrical, it takes 2.0 s for the ball to reach the highest point and then another 2.0 s to land back on the tee.

Therefore, we can just analyze the second half of the motion that lasts

t = 2.0 s

During this time, the vertical distance covered by the ball is given by the equation:

$d=ut+\frac{1}{2}gt^2$

where

u = 0 is the initial velocity (zero because the ball starts from its highest point, where the velocity is zero)

t = 2.0 s

g = 9.8 m/s^2 is the acceleration of gravity

Solving for d, we find:

$d=\frac{1}{2}(9.8)(2.0)^2=19.6 m$

So, the ball reaches a maximum height of 19.6 m.