Problema Solution
An object is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t2+ 19.6t + 58.8, where s is in meters.
What was its maximum height?
Answer provided by our tutors
An object is launched from a height so it follows projectile path so for this at maximum height means where the final velocity is =0.
Given Initial velocity u=19.6 m/s
final velocity according to theory=0m/s
Here the distance is not needed so from projectile motion theory max height h=((u^2)/(2*g))
Here u=initial velocity=19.6m/s
g=gravity constant=9.8m/s (for earth)
so max height=h= (u^2)/(2*g)= (19.6*19.6)/(2*9.8)=19.6 meters
So maximum height reached by the launched body = 19.6 meters..