A0.060 kg ball hits the ground with a speed of -32m/s. the ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 n force on the ball. what is the magnitude of the velocity after it hits the ground?

a. 9.3m/sb.12m/sc.41m/sd.73m/s

A:

Impulse is given by:

FΔt = Δp

We know the mass and velocity and we need the final velocity so lets break this down and isolate

FΔt = mv(final) - mv(initial)

Factor out m

FΔt = m(v(final) - v(initial))

FΔt / m = v(final) - v(initial)

Add v initial

(FΔt / m) + v(initial) = v(final)

(55x45ms / 0.060) - 32 = v(final)

v(final) = +9.25 m/s or 9.3

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

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explanation: