Respuesta :
Answer:
1) h(max) = 64 ft
2) t = 1 s
3) t(in air) = 3 s
4) just after 3 s
Step-by-step 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)
