A projectile is thrown upward so that it's distance above the ground after t seconds is h=-16t^2+480t. After how many seconds does it reach its maximum height?

Question

asked Feb 1, 2021 15.1k views

1 Answer

Matti Mehtonen · Answer 1 · 2021-02-05T17:53:01+0000

Answer:

t = 15 s

Explanation:

A projectile is thrown upward so that it's distance above the ground after t seconds is as follows :

h=-16t^2+480t ...(1)

We need to find after how many seconds does it reach its maximum height. For maximum height, put
(dh)/(dt)=0

$(d(-16t^2+480t))/(dt)=0\\\\-32t+480=0\\\\t=(480)/(32)\\\\t=15\ s$

So, it will reach its maximum height in 15 seconds.