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

Step-by-step 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. $\frac{ds}{dt}=0$

$\frac{ds}{dt}=-14t+14+0=0$

$t=\frac{14}{14}=1\ s$

Therefore after 1 sec object achieves maximum height

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

$s(1)=343\ m$