Question #a0334

1 Answer
Jun 8, 2017

Answer:

#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.