Answer to Question #241459 in Molecular Physics | Thermodynamics for Kelani

Question

Pedro and Joanna decide to try a problem.

A baseball is thrown from the top of a tall building with an initial velocity of 10.5 m/s from a height of h = 10 m above the ground.

Find its speed when it reaches the ground if its launch angle is 37°.

v_f =___  m/s

Find its speed if it is launched horizontally.

v_f = ___ m/s

Answer 1

In the first case, we will use the following equations:

$h_\text{up}=\frac{v^2\sin^2\theta}{2g},\\\space\\ h_\text{down}=h_\text{up}+h=\frac{v^2\sin^2\theta}{2g}+h=12\text{ m}.$

The vertical component of velocity:

$v_y=\sqrt{2gh_\text{down}}=15.4\text{ m/s},\\ v_x=v\cos\theta=8.39\text{ m/s}^2.\\\space\\ v=\sqrt{v_y^2+v_x^2}=17.5\text{ m/s}.$

In the second case:

$v_y=\sqrt{2gh}=14\text{ m/s},\\ v_x=v\cos\theta=8.39\text{ m/s}^2.\\\space\\ v=\sqrt{v_y^2+v_x^2}=16.3\text{ m/s}.$