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:
$K= \frac{1}{2} mv^2$
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
$\frac{1}{2}mv^2=mgh$
which means
$v= \sqrt{2gh}$
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:
$v= \sqrt{2gh}= \sqrt{2(9.81 m/s^2)(45 m)}=29.7 m/s$

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
$h= \frac{1}{2} gt^2$
where t is the time the ball takes to reach the altitude h=45 m. Rearranging the equation, we find
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(45 m)}{9.81 m/s^2} }=3.02 s$