Aboy is to catch a ball which drops vertically from the top of a 75.1 m tower. find the average velocity at whihc the boy, initially 33.4 m from the base of the tower, must run to be able to catch the ball just before it hits the ground

boy average velocity to catch the ball just before it hits the ground is 8.53 m/s

Explanation:

given data

vertical distance = 75.1 m

horizontal distance = 33.4 m

to find out

Find the average velocity to catch the ball just before it hits the ground

solution

we know here ball is coming downward and boy is running to catch it

so first we calculate the time of ball to reach at ground that is express by equation of motion

s = ut + 0.5 × at²    .....................1

here u is initial speed that is zero and a = 9.8 and s is distance

put here all value to get time t  in equation 1

75.1 = 0 + 0.5 × (9.8)t²  

t = 3.914 s

so

boy speed is =

speed =

speed = 8.53 m/s

so average velocity to catch the ball just before it hits the ground is 8.53 m/s

we need to know the time interval during which the force acts. impulse of force = momentum change

overcharge is the right answer system