Question

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

Answer 1

Two missing questions:

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

Express your answer to two significant figures and include the appropriate units.

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

Express your answer using two significant figures."

A) 1500 N

First of all, we have to calculate the velocity at which the impala jumps upward. This can be done by using the motion equation:

$v^2-u^2 = 2ah$

where

v = 0 is the velocity of the impala as it reaches the maximum height

u is the initial velocity of the impala

a = g = -9.8 m/s^2 is the acceleration of gravity

h = 2.5 m is the maximum height reached

Solving  for u,

$u=\sqrt{-2ah}=\sqrt{-2(-9.8)(2.5)}=7 m/s$

Now we know that the impulse on the impala is equal to its change in momentum, so we can write:

$I=\Delta p \rightarrow F \Delta t = m \Delta v$

where

F is the force exerted by the ground on the impala

$\Delta t = 0.21 s$ is the duration of the push

m = 45 kg is the mass

$\Delta v = 7 m/s$ is the gain in velocity of the impala

Solving for F,

$F=\frac{m\Delta v}{\Delta t}=\frac{(45)(7)}{0.21}=1500 N$

And according to Newton's third law, the force exerted by the impala on the ground is equal (and opposite) to the force exerted by the ground on the impala, so still 1500 N.

B) 3.4

The weight of the impala is given by

$W=mg$

where

m = 45 kg is its mass

g = 9.8 m/s^2 is the acceleration of gravity

Substituting,

$W=(45)(9.8)=441 N$

So the ratio of the force calculated at point A) to the weight of the impala is

$r=\frac{F}{W}=\frac{1500}{441}=3.4$