Question

A baseball is thrown straight up in the air with an

initial velocity of 128 feet per second from a point
exactly 4 feet off the ground.
Write the function for the height of this object at any
time (t seconds)

Answer 1

Explanation:

`h(t)=-16t^2+v_0t+h_0` where

h(t) is the height of the ball after a certain amount of time has gone by,

-16t² is the pull of gravity in ft/s/s,

v₀t is the initial upwards velocity, and

h₀ is the height from which the ball was initially thrown. Our equation is then, filling in the info in the correct places:

`h(t)=-16t^2+128t+4` where 128 is the given initial velocity and 4 is the number of feet from which the ball was initially launched.