Question

A heavy red ball is released from rest 2.0 m above a flat, horizontal surface. At exactly the same instant, a yellow ball with the same mass is fired horizontally at 3.0 m/s. Which ball hits the ground first?

Answer 1

Answer:

Both ball hit the ground about at the same time.

Explanation:

given information:

h = 2 m

the speed of red ball, v = 3 m/s

the time for red ball to reach the ground

h = $\frac{1}{2}gt^{2}$

t = √2h/g

 = √2(2)/9.8

 = 0.64s

the time for yellow ball to reach the ground, it's considered as a vertical motion. thus

h = √2h/g

 = √2(2)/9.8

 = 0.64s

so, both ball hit the ground about at the same time.