Question

A ball is thrown horizontally from the top of a 64.0 m building and lands 106.0 m from the base of the building. Ignore air resistance.
(a) How long is the ball in the air? s
(b) What must have been the initial horizontal component of the velocity? m/s
(c) What is the vertical component of the velocity just before the ball hits the ground? m/s
(d) What is the horizontal component of the velocity of the ball just before it hits the ground? m/s

Answer 1

Let's use SUVAT:
We know that the initial velocity is zero.
We know that the building is 64 m high.
We know that it lands 106.0 m from the base:

a)
The forces is the y-direction (i.e. vertical) are all dependent on gravity:
$$
s=ut-\cfrac{1}{2}gt^2\\
t=\sqrt{\cfrac{2s}{g}}=\sqrt{\cfrac{2\times 64}{9.81}}~seconds
$$
b)
Their is no horizontal acceleration:
$$
s=ut-\cfrac{1}{2}gt^2\\
=ut\\
u=\cfrac{s}{t}
$$
We know \(s\) (i.e 106.0 m) and we know \(t\) (see part a) so we know \(u\), our starting horizontal velocity.
c)
The vertical component is completely independent of the initial horizontal velocity and only depends on gravity:
$$
v^2=u^2-2gs\\
v=\sqrt{-2\times9.81\times-64}
$$
d)
There are no horizontal forces, so the final horizontal velocity is equal to the initial horizontal velocity (determined in part b above).

That's it!