Thread: Height of a ball

The height of a ball (in feet) t seconds after it is thrown is given by h(t)= -16t^2+ 38t + 74; its (upward) velocity at time t is v(t)= -32t + 38.

(a) From what height was the ball thrown?
(b) What was the ball's initial velocity? Was it thrown up or down? How can you tell?
(c) Was the ball's height increasing or decreasing at time t=2?
(d) At what time did the ball reach its maximum height? How high was it then?
(e) How long was the ball in the air?

My answers

(a) 96 feet (I plotted it on a graph)
(b) 74mph, the ball was thrown up because the velocity is a large number
(c) The ball's height was increasing
(d) At t=1 it was 96 feet
(e) 6 minute

Are these correct?

No,

a)You have the equation for height, so the initial height of the ball would be the height at time 0, plug in t=0

b)You also have an equation for velocity, so for initial velocity plug in t=0

For c) plug 2 into the equation for velocity, if velocity is negative, then the height is decreasing

For d) you can either use what you know about the parabola, or basic physics would tell you that at the max height, the balls velocity is 0, so solve when v(t)=0

For e) solve for when the height =0, and get the time...

how did you get your answers to begin with?

a. You want h(0) = 74

b. v(0) = 38

c. v(2) = -26 means its height was decreasing

d when it reaches its max height v(t) = 0

t= 38/32 = 19/16

e Set h = 0 -16t^2+ 38t + 74 = 0 t = 3.64 secs usinq quadratic formula