Question

A 70.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 15.2 inches off the ground. He hits the ground 1.11 inches away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?

Answer 1

initial vertical height, h = 15.2 inch

= 15.2*2.54*10^-2

= 0.386 m

x = 1.11 inch

= 1.11*2.54*10^-2 m

= 0.028194 m

let t is the time taken for player to strike the groud.

use,

y = voy*t + 0.5*ay*t^2

h = 0 + (1/2)*g*t^2

==> t = sqrt(2*h/g)

= sqrt(2*0.386/9.8)

= 0.28067 s

let V is the velocity of the player after catching the ball.

so, V = x/t

= 0.028194/0.28067

= 0.10045 m/s

let u is the initial speed of the ball before cacthing.

now use conservation of momentum

m*u = (m + M)*V

==> u = (m + M)*V/m

= (0.43 + 70)*0.10045/0.43

= 16.4 m/s <<<<<<<<<<-------------------Answer