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

+ 476t. what is the maximum height?

y (max) = 4046 m

Step-by-step 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