Question

A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 142 feet per second from an initial height of 93 feet off the ground, then the height of the projectile, h h, in feet, t t seconds after it's shot is given by the equation:

Answer 1

Explanation:

It is given that,

Initial velocity of the projectile, u = 142 ft/s

Initial height off the ground, $h_o = 93\ feet$

We need to find the height of the projectile t seconds after its shot. It is a concept of kinematics. The equation of projectile is given by the formula as follows :

$h= -16t^2+ut+h_o$

t is time in seconds

So, putting all the values we get :

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

Hence, this is the required solution.