Question

The equation h(t)=−16t2+25t+80 gives the height of a ball, in feet, t seconds after it is thrown from a platform.
What is the initial velocity when the ball is thrown?

Answer 1

take the derivitive

h'(t)=-32t+25
initial is at t=0
h'(0)=-32(0)+25
h'(0)=0+25
h'(0)=25

initial velocity is 25ft/sec

Answer 2

Answer:

initial velocity when the ball is thrown is 25 ft/sec.

Step-by-step explanation:

The height of the object of the projectile motion is given by:

$h(t) = -at^2+v_0t+h_0$           ....[1]

where

a is the acceleration due to gravity i.,e a = 16 ft/s^2

$v_0$ is the initial velocity

$h_0$ is the initial height of the object.

Given the equation:

$h(t) = -16t^2+25t+80$

where

h(t) is the height of the ball in feet after it is thrown from a platform.

t is the time in seconds

On comparing the given equations with [1] we have;

⇒$v_0 = 25$ ft/sec

therefore, the initial velocity when the ball is thrown is 25 ft/sec.