Question

A projectile is thrown upward so that its distance is h=14t square

+ 476t. what is the maximum height?

Answer 1

y (max) = 4046 m

Explanation:

h = 14*t² + 476*t is not the equation for a projectile upward movemente, since in that case projectile will continue going up to nfinite as time goes on. We must correct the equation according to projectile shot. Then

y = h = - 14*t² + 476 * t (1)

So we will have h maximum when V(y) = 0

then D(y)/dt = - 28 *t + 476

D(y)/dt = 0 ⇒ - 28 *t + 476 = 0 ⇒ t = 476 / 28

t = 17 sec

Then y maximum is for t = 17 plugging this value in equation (1) we get

y = - 14*t² + 476* t

y (max) = - 14* (17)² + 476 * ( 17) ⇒ y (max) = - 4046 + 8092

y (max) = 4046 m