HW QUESTION-I HAVE BEEN STUMPED FOR DAYS! PLEASE HELP DUE TONIGHT!

Question

An impala is an African antelope capable of a remarkable vertical leap. In one recorded leap, a 45 kgkg impala went into a deep crouch, pushed straight up for 0.21 ss, and reached a height of 2.5 mm above the ground.

To achieve this vertical leap, with what force did the impala push down on the ground?

What is the ratio of this force to the antelope's weight?

NOTE:

(the answer for the first question is NOT 2550 N, 441 N , 220.5 N, 265.5 N, 486 N)

(the answer to the second question is NOT 56.73.4, 33.3)

Answer 1

If the height attained was 2.5 m, we can use the kinematic equation:

v_f² = v_i² + 2ax

to find the initial velocity (v_i). To do so, we use v_f = zero, a = -9.8 m/s², and x = 2.5 m:

0² = v_i² + 2(-9.8)(2.5)

0 = v_i² - 49

v_i = 7 m/s

So a force was used to change the vertical velocity from zero (on the ground prior to the jump) to 7 m/s (just slightly above the ground after the force was expended), and that velocity was enough to lift the impala to a height of 2.5 m.

The impulse-momentum theorem say the impulse equals the change in momentum.

The impulse = F•t = F(0.21)

The change in momentum = mv₂ - mv₁ = m(v₂ - v₁) = 45(7 - 0) = 315 kgm/s

And, since impulse equals the change in momentum, then F(0.21) = 315 so F = 315/0.21 = 1500 N

To get the ratio of force to weight, first determine the weight. W = mg = (9.8)(45) = 441 N

Ratio of force/weight = 1500/441 = 3.4