Question

Question #a0334

Answer 1

`21.9` `"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

`(v_y)^2 = (v_(0y))^2 +2a_y(y - y_0)`

to find this height. We have

• `v_y`is `0` (this is when it's at its maximum height)

• `v_(0y)` is `20"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` `"m"`

Plugging in known variables, we have

`0 = (20"m"/"s")^2 -2(9.8"m"/("s"^2))(y - 1.5"m")`

`(19.6"m"/("s"^2))(y-1.5"m") = 400("m"^2)/("s"^2)`

`(19.6"m"/("s"^2))(y) - 29.4("m"^2)/("s"^2) = 400("m"^2)/("s"^2)`

`(19.6"m"/("s"^2))(y) = 429.4("m"^2)/("s"^2)`

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

Thus, the maximum height it reaches is `21.9` `"m"` above the ground.