Question

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

a. what must be the balls 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 balls maximum height. show all of your work below.

1. a ball is thrown straight up with an initial velocity of \v0\. it takes the ball \t\ seconds to reach its maximum height. how long must it have taken for the ball to reach half of its initial velocity v0/2? answer in terms of t and explain your answer briefly.

Explanation:

Step1: Analyze velocity at max height

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

Step2: Find time to reach max height

Use the 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). Rearranging for $t$ gives $t=\frac{v - v_0}{a}$. Substituting the values: $t=\frac{0 - 14}{-9.8}=\frac{14}{9.8}\approx1.43$ s.

Step3: Calculate max height

Use the 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 $h=\frac{v^{2}-v_{0}^{2}}{2a}$. Substituting the values: $h=\frac{0 - 14^{2}}{2\times(-9.8)}=\frac{- 196}{-19.6}=10$ m.

Step4: Solve for time to reach half - initial velocity

We know that $v = v_0+at$. We want to find $t$ when $v=\frac{v_0}{2}$. Substituting into the equation gives $\frac{v_0}{2}=v_0+at$. Rearranging for $t$: $at=\frac{v_0}{2}-v_0=-\frac{v_0}{2}$, so $t = \frac{-\frac{v_0}{2}}{a}$. We also know that at max - height $v = 0$ and $v = v_0+at_{max}$, so $0 = v_0+at_{max}$, and $t_{max}=\frac{-v_0}{a}$. Since $t_{max}=t$ (time to reach max height), and $t_{half - v_0}=\frac{-\frac{v_0}{2}}{a}$, we can see that $t_{half - v_0}=\frac{t}{2}$.

Answer:

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

1. $\frac{t}{2}$