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

42.28 m/s

Step-by-step explanation:

Answer 2

v = 42.28 m/s

Step-by-step 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+(1)/(2)at^2`

a = -g and u = 0

`d=-(1)/(2)gt^2\\\\t=\sqrt{(-2d)/(g)} \\\\t=\sqrt{(-2* 1.25)/(9.8)} \\\\t=0.5\ s`

Vertical component of the velocity is :

`v_y=gt\\\\v_y=9.8* 0.5\\\\v_y=4.9\ m/s`

Resultant velocity,

`v=√(v_x^2+v_y^2) \\\\v=√(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.