a projective is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet. If height after t seconds is represented by h(t) = -16t^2 + 64t + 96, find the maximum height the projective reaches

Answer 1

To find the maximum height the projectile reaches, we need to find the vertex of the parabolic function h(t) = -16t^2 + 64t + 96.

The vertex of a parabolic function in the form h(t) = at^2 + bt + c is given by the formula t = -b/(2a).

In this case, a = -16 and b = 64. So the time at which the maximum height is reached is t = -64/(2*(-16)) = -64/(-32) = 2 seconds.

To find the maximum height reached by the projectile, we substitute t = 2 into the equation h(t):

h(2) = -16(2)^2 + 64(2) + 96
h(2) = -16(4) + 128 + 96
h(2) = -64 + 128 + 96
h(2) = 160

Therefore, the maximum height the projectile reaches is 160 feet.