A baseball player throws a 95mph fastball straight up into the air. The position equation, which gives the height of the ball at

Question

3 years ago

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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!

Mathematics

1 answer:

Lana71 [14] · Answer 1 · 2022-06-08T08:30:34+03:00

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.
<span>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
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