Question

A projectile is thrown upward so that its distance above the ground after t seconds is h=-121t^2 +504t. After how many seconds does it reach it maximum height?

Answer 1

Answer:

2.08secs

Step-by-step explanation:

Given the equation modeled by the height as h=-121t^2 +504t

The velocity of the object at the max height is zero, hence;

v = dh/dt = 0

dh/dt = -2(121)t + 504

dh/dt = -242t + 504

0 = -242t + 504

242t = 504

t = 504/242

t = 2.08secs

Hence the projectile reaches its maximum height after 2.08secs