Question

classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per second from an initial height of 112 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= -16t^2+128t+112 Find the two points in time when the object is 147 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).

Answer 1

`t_1 = 0.28 s`

`t_2 = 7.72 s`

Step-by-step explanation:

Given that height of the projectile as a function of time is

`h = -16 t^2 + 128 t + 112`

here we know that

h = 147 ft

so from above equation

`147 = -16 t^2 + 128 t + 112`

`16 t^2 - 128 t + 35 = 0`

now by solving above quadratic equation we know that

`t_1 = 0.28 s`

`t_2 = 7.72 s`