A ball is thrown upward with a velocity of 117 m/sec and it started off at a height of 38 m. It is modeled by the equation h(t) = - 4.9t^2 + 117t + 38 where the height of the ball is h(t) after time t (in seconds). What is the maximum height the ball achieves?

252 m

155 m

12 m

736 m

Question

A ball is thrown upward with a velocity of 117 m/sec and it started off at a height of 38 m.  It is modeled by the equation h(t) = - 4.9t^2 + 117t + 38 where the height of the ball is h(t) after time t (in seconds).  What is the maximum height the ball achieves?

		
252 m

		
155 m

		
12 m

		
736 m

pratyush5 · Answer

You know calculus ?

pratyush5 · Answer

BTW I did something and my answer didn't match. Maybe its my mistake. I'll try again.

pratyush5 · Answer

It matched. So what I basically did - 

1. Diffrentiate the height with respect to time. We get an expression for velocity. 
2. At maximum height velocity will be zero, so equate that expression with zero.
3. We get value of time at which ball will be at max height
4. Input tht value of time into the equation of height given.
5. You'll get the answer :)

pratyush5 · Answer

Shall I explain it step wise ?

anonymous · Answer

Refer to the Mathematica attachment.