Question

A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 151 feet per second from an initial height of 95 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation:
h = - 16t² + 151t +95

Answer 1

Answer:

Step-by-step explanation:

144 = -16t² + 151t + 95

  0 = -16t² + 151t - 49

quadratic formula

t = (-151 ±√(151² - 4(-16)(-49))) / (2(-16))

t = (-151 ± 140.23195) / -32

t = 0.34 s

t = 9.10 s