Question

A student throws a softball horizontally from a dorm window 15.0 m above the ground. Another student standing 10.0 m away catches the ball at a height of 1.50 m above the ground. What is the initial speed of the ball?

Answer 1

Answer:

Speed of ball equals 6.024 m/s.

Explanation:

Let the student throw the ball with a velocity of 'v' m/s horizontally.

Now the time in which the ball travels 10.0 meter horizontally shall be equal to the time in which it travels (15.0-1.50) meters vertically

Hence the time taken to cover a vertical distance of 13.50 meters is obatined using 2 equation of kinematics as

$s=\frac{1}{2}gt^{2}\\\\t=\sqrt{\frac{2s}{g}}\\\\t=\sqrt{\frac{2\times 13.5}{9.81}}\\\\\therefore t=1.66 seconds$

Since there is no acceleration in horizantal direction we infer that in time of 1.66 seconds the ball travels a distance of 10 meters

Hence the spped of throw is obatines as

$Speed=\frac{Distance}{Time}\\\\v=\frac{10}{1.66}=6.024m/s$