Question

1. a batter hits a ball straight up with an initial velocity of 14 m/s.

a. what must be the ball’s velocity when it reaches its maximum height?
b. how long does it take the ball to reach its maximum height? show all of your work below.
c. calculate the ball’s maximum height. show all of your work below.

Explanation:

Step1: Analyze velocity at max - height

At the maximum height of a vertically - thrown object, its velocity is 0 m/s because the object momentarily stops before starting to fall back down.

Step2: Calculate time to reach max - height

Use the kinematic equation $v = v_0+at$. Here, $v = 0$ (velocity at max - height), $v_0 = 14$ m/s (initial velocity), and $a=-g=- 9.8$ m/s² (acceleration due to gravity, negative as it acts in the opposite direction of motion). Rearranging for $t$ gives $t=\frac{v - v_0}{a}$.
$t=\frac{0 - 14}{-9.8}=\frac{-14}{-9.8}\approx1.43$ s

Step3: Calculate maximum height

Use the kinematic equation $v^{2}=v_{0}^{2}+2ah$. At max - height $v = 0$, $v_0 = 14$ m/s, and $a=-9.8$ m/s². Rearranging for $h$ gives $h=\frac{v^{2}-v_{0}^{2}}{2a}$.
$h=\frac{0 - 14^{2}}{2\times(-9.8)}=\frac{- 196}{-19.6}=10$ m

Answer:

a. 0 m/s
b. Approximately 1.43 s
c. 10 m