Question

A pitcher throws a baseball horizontally with a speed of 42 m/s. If the pitcher throws the baseball from a height of 1.25 meters, what is the resultant velocity of the ball right before it hits the ground? Include direction (angle).

Answer 1

Answer:

v = 42.28 m/s

Explanation:

It is given that,

The horizontal speed of a baseball, $u_x=42\ m/s$

The pitcher throws the baseball from a height of 1.25 meters, h = 1.25 m

Firstly finding t i.e. the stone takes to reach the ground. Using second equation of motion as follows :

$d=ut+\dfrac{1}{2}at^2$

a = -g and u = 0

$d=-\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{-2d}{g}} \\\\t=\sqrt{\dfrac{-2\times 1.25}{9.8}} \\\\t=0.5\ s$

Vertical component of the velocity is :

$v_y=gt\\\\v_y=9.8\times 0.5\\\\v_y=4.9\ m/s$

Resultant velocity,

$v=\sqrt{v_x^2+v_y^2} \\\\v=\sqrt{42^2+4.9^2} \\\\v=42.28\ m/s$

So, the resultant velocity of the ball before it hits the ground is 42.28 m/s.

Answer 2

Answer:

42.28 m/s

Explanation: