Question

A rock is thrown straight up into the air with an initial velocity of 44ft/s. The height of the rock after t seconds is expressed h(t) = -16t^2 + 44t.

1.) How long does it take until the ball hits the ground?

2.) What is the maximum height the ball reaches?

Answer 1

1.

When rock reaches the ground height becomes zero.

0 = h(t) = -16t^2 + 44t.

16t^2 = 44t

t = 44/16 = 2.75 seconds

Answer 2

2.

v^2=u^2+2as. 

u = initial velocity = 44 ft/s

v=final velocity = 0

s = maximum geight = h 

a= -32.174 ft/s^2 (Deceleration due to gravity) 

Substitute the known values

0 = 44^2 + 2*(-32.174) * h (ft)

64.348 h = 1936

h = 30.086 ft