Question

Crystal kick the 3.0 kg soccer ball horizontally off of the top of a building the building is 6.0 m tall and the ball lands 10.0 m away from the base of the building what was the bus initial velocity

Answer 1

The ball initial velocity is 9.1 m/s

Explanation:

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path, which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

First of all, we analyze its vertical motion to find the time of flight. We can do it by using the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$ (1)

where :

$u_y = 0$ is the initial vertical velocity  (0 because the ball is kicked horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -6.0 m is the vertical displacement (the height of the building)

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(-6.0)}{-9.8}}=1.1 s$

Now we can analyze the horizontal motion; the horizontal velocity is constant and it is given by

$v_x = \frac{d}{t}$

where

d = 10.0 m is the horizontal distance covered

t = 1.1 s is the time of flight

Substituting, we find

$v_x = \frac{10.0}{1.1}=9.1 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly