Question #a0334

1 Answer
Nathan L.
Jun 8, 2017

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_yis 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.

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