Question

An object is launched at 28 meters per second (m/s) from a 336-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –7t^2 + 14t + 336 where s is in meters

What is it looking for when the question asks "What is the maximum height?"

Answer 1

Answer:343 m

Explanation:

Given

launch velocity of object is
`u=28\ m/s`

height of Platform
`h=336\ m`

height of object is given by

`s(t)=-7t^2+14t+336`

For maximum height velocity of object is zero

i.e.
`(ds)/(dt)=0`

`(ds)/(dt)=-14t+14+0=0`

`t=(14)/(14)=1\ s`

Therefore after 1 sec object achieves maximum height

`s(1)=-7(1)+14(1)+336`

`s(1)=343\ m`