Question

The equation h(t)=-16t^2+32t+12 gives the height of a baseball, in feet, t seconds after it is thrown from a platform. What is the height of the platform? What is the initial velocity when the baseball is thrown?

Answer 1

Answer:

(a) 12 m

(b) 32 m/s

Step-by-step explanation:

(a) The height of the platform is h(0) i.e. the height, h, at time t = 0 secs, since the ball would not have been thrown at that time.

Therefore, h(0) is:

$h(0) = -16(0^2) + 32(0) + 12\\\\\\h(0) = 0 + 0 + 12\\\\\\h(0) = 12 m$

The height of the platform is 12 m.

(b) The initial velocity when the baseball is thrown will be v(0) that is velocity when t = 0 secs.

We obtain velocity, v, by differentiating height, h, with respect to time:

$v(t) = \frac{dh}{dt} = -32t + 32$

Therefore, at time t = 0 secs:

$v(0) = -32(0) + 32\\\\\\v(0) = 32 m/s$

The initial velocity of the baseball when it is thrown is 32 m/s.