A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer player's foot at a velocity of 8.50 m/s. If the time of contact between the ball and the player's foot
was 2.00 x 10-2 seconds, what was the force that the foot applied on the ball?

this is the second question

A 1.50 kg rock falls from the top of a 10.0 m high building and strikes the ground below. Calculate
the impulse experienced by the rock during its fall.​

A)   F = - 8.5 10² N,  B)   I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

          I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

         v₀ = 8.50 m / s

         v_f = -8.50 m / s

         F t = m v_f -m v₀

         F =

let's calculate

         F =

         F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

         v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

         v =

calculate  

         v =

         v = 14 m / s

to calculate the momentum we use

         I = Δp

         I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

         I = 1.50 (0-14)

         I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

        I = 21 N s

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )