Question

PLEASE HELP ASAP!!!! A ball is thrown up in the air from the top of a 48 ft. building. The height of the ball, h(t), after t seconds is given by the equation h(t) = -16t2 + 32t + 48 1) What is the maximum height of the ball? 2) At what time does the ball reach the maximum height? 3) For how long is the ball above the ground? 4) When does the ball hit the ground? Explain your solutions for this context.

Answer 1

1) h(max) = 64 ft

2) t = 1 s

3) t(in air) = 3 s

4) just after 3 s

Explanation:

The height f the ball is:

h(t) = -16*t² + 32*t + 48

Maximum height is when V(y) = 0

D(h)/dt = V(y) = -16*2*t + 32

V(y) = 0 -32*t + 32 = 0

2) t = 1 s time to reach maximum height

Plugging this value in equation h(t) = -16*t² + 32*t + 48

h(max) = -16*(1)² + 32*(1) + 48

1) h(max) = 16 + 48 = 64 ft

3) In equation h(t) = -16*t² + 32*t + 48

when h(t) = 0 we find time of the ball in air

16*t² - 32*t - 48 = 0 a second degree equation solving for x

x₁,₂ = 32 ± √( (32)² + 4*48*16) / 32

x₁,₂ = (32 ± 64 )/32

we dismiss negative x ( negative time)

x₂ = 32 + 64 /32

x₂ = 3 s

4) the ball hit the ground just at 3 scond after its thrown

The ball is 3 s in the air ( touch the ground just in that moment)