Question a0334

Jun 8, 2017

$21.9$ $\text{m}$

Explanation:

I'll assume the maximum height is how far the baseball reaches above the ground, rather than the pitcher.

We need to find the height $y$ when it reaches its maximum height. A projectile is at its maximum height when its instantaneous velocity is zero. We can use the equation

${\left({v}_{y}\right)}^{2} = {\left({v}_{0 y}\right)}^{2} + 2 {a}_{y} \left(y - {y}_{0}\right)$

to find this height. We have

• ${v}_{y}$is $0$ (this is when it's at its maximum height)

• ${v}_{0 y}$ is $20 \text{m"/"s}$

• ${a}_{y}$ is $- g$, which is -9.8"m"/("s"^2)

• $y$ is the height we wan to find, and

• ${y}_{0}$ is its initial height, $1.5$ $\text{m}$

Plugging in known variables, we have

$0 = \left(20 \text{m"/"s")^2 -2(9.8"m"/("s"^2))(y - 1.5"m}\right)$

$\left(19.6 {\text{m"/("s"^2))(y-1.5"m") = 400("m"^2)/("s}}^{2}\right)$

$\left(19.6 {\text{m"/("s"^2))(y) - 29.4("m"^2)/("s"^2) = 400("m"^2)/("s}}^{2}\right)$

$\left(19.6 {\text{m"/("s"^2))(y) = 429.4("m"^2)/("s}}^{2}\right)$

y = color(red)(21.9 color(red)("m"#

Thus, the maximum height it reaches is $21.9$ $\text{m}$ above the ground.