Question

A ball is thrown vertically up with a velocity of 80 ft/sec at the edge of a 200-ft cliff. Calculate the height h to which the ball rises and the total time t after release for the ball to reach the bottom of the cliff. Neglect air resistance and take the downward acceleration to be 32.2 ft/sec.​

Answer 1

y = ut +
`(1)/(2)`
`at^(2)` , y = 80t -
`(1)/(2)`
`32.2t^(2)`

for y = -200 ft,

-200 = 80t -
`16.1t^(2)` or
`16.1t^(2)` - 80t - 200 = 0

t = 80 ±
`\frac{\sqrt{80^(2) + 4(16.1)(200) }}{2(16.1)}` = 6.80 sec. ( or - 1.83 s)

For y = 0,
`u^(2)` =
`u_(o)^(2)` + 2ay, y = h =
`(0-80^(2) )/(-2(32.2))`

= 99.4 ft.

Step-by-step explanation:

Hope this helped, please message me if I made a mistake!

-Matt