Question

You decide to launch a ball vertically so that a friend located 45 m above you can catch it. what is the minimum launch speed you can use? how long after the ball is launched will your friend catch it?

Answer 1

1) We can solve this part of the problem by using the law of conservation of energy. In fact, the initial energy of the ball is just kinetic energy:

`K= (1)/(2) mv^2`
where m is the mass of the ball and v is the initial speed of the ball. At its maximum height, the speed of the ball is zero, so its energy is just gravitational potential energy:

`U=mgh`
where g is the gravitational acceleration and h is the height above the ground.
Since the total energy must be conserved, we have

`(1)/(2)mv^2=mgh`
which means

`v= √(2gh)`
If we use h=45 m, we find the minimum speed v such that the ball reaches an altitude of 45 m above the ground:

`v= √(2gh)= √(2(9.81 m/s^2)(45 m))=29.7 m/s`

2) The motion of the ball is an accelerated motion, and the relationship between the distance covered by the ball and the time t is given by

`h= (1)/(2) gt^2`
where t is the time the ball takes to reach the altitude h=45 m. Rearranging the equation, we find

`t= \sqrt{ (2h)/(g) }= \sqrt{ (2(45 m))/(9.81 m/s^2) }=3.02 s`