A baseball player throws a 95mph fastball straight up into the air. The position equation, which gives the height of the ball at any time t, in seconds, is given by: s (t) = - 16t^2+140t+37 a.) find the maximum height of the ball. b.) Find the time and velocity when the ball hits the ground.

I would really appreciate it if someone explained to me how to solve this!

Question

Mathematics

Damari Byrd

2 April, 01:28

A baseball player throws a 95mph fastball straight up into the air. The position equation, which gives the height of the ball at any time t, in seconds, is given by: s (t) = - 16t^2+140t+37 a.) find the maximum height of the ball. b.) Find the time and velocity when the ball hits the ground.

I would really appreciate it if someone explained to me how to solve this!

+2

Answers (1)

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Regan Chan · Answer 1

Set s (t) = 0 which means it hits the ground. The formula doesn't fit the parameters given as it shows that the pitcher is standing on something 37 feet high.

s (t) = - 16t^2+140t+37 and has an initial velocity of 140.

Graphing or solving this, t = 9 seconds when it hits the ground.

The velocity V (t) is the derivative of s (t)

V (t) = - 32t+140

V (9) = - 148 ft/second which is going down