Question

a ball is thrown upward from a bridge with a speed of 48 ft/s. It misses the bridge on way down and lands in water 160 ft. below. Find how long the ball rises, how high it goes, how long it is in the air, and its velocity when it strikes the water?

Answer 1

h(t) = 160 + 48t - 16t^2
v(t) = 48 - 32t
See what you can do with that.

Answer 2

1. V = Vo + g*Tr = 0.,
48 + (-32)Tr = 0,
Tr36 + = 1.5 s. = Rise time.

2. V^2 = Vo^2 + 2g*h = 0.
48^2 + (-64)h = 0,
h = 36 ft.

3. h = 0.5g*Tf^2 = 36 + 160.
16*Tf^2 = 196,
Tf = 3.5 s. = Fall time.
Tr + Tf = 1.5 + 3.5 = 5.0 s. = Time in air.

4. V = Vo + g*Tf = 0 + 32*3.5 =