Question

A ball is thrown vertically up with a velocity of 65 ft/sec at the edge of a 680-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/sec2.

Answer 1

a)
`y=y_(0)+v_(oy) t+(1)/(2) gt^(2)  =640 ft+(65ft/s)(2.018s)+((1)/(2))(-32.2ft/s^(2) )(2.018s)^(2)`

`y=705.6ft`

b)
`t=8.63 s`

Step-by-step explanation:

We start the exercise knowing that a ball is thrown up with an initial velocity of 65 ft/s with an initial height of 680 ft.

To calculate the maximum heigh, we know that at the top of the motion the ball stop going up and start going down because of the gravity

First of all, we need to calculate the time that takes the ball to reach the maximum point.

a)
`v_(y)=v_(oy)+gt`

`t=(-v_(oy) )/(g)=(-65ft/s)/(-32.2ft/s^(2) ) =2.018s`

Knowing that time, we can calculate the height to which the ball rises:

`y=y_(0)+v_(oy) t+(1)/(2) gt^(2)  =640 ft+(65ft/s)(2.018s)+((1)/(2))(-32.2ft/s^(2) )(2.018s)^(2)`

`y=705.6ft`

b) Now, to know the time that the ball reach the bottom of the cliff, we know that the final height is y=0ft

`y=y_(0)+v_(oy) t+(1)/(2) gt^(2)`

`0=640ft+(65ft/s)t-(1)/(2)(32.2ft/s^(2))t^(2)`

This is a classic quadratic equation, that can be solve using the quadratic formula

`t=\frac{-b±\sqrt{b^(2)-4ac } }{2a}`

a=-16.1

b=65

c=640

Solving for t, we have that

`t=-4.6015 s` or
`t=8.6387s`

Since the time can not be negative:

`t=8.6387s`