An impala is an African antelope capable of a remarkable vertical leaf. 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. 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? What is the ratio of this force to the antelope's weight?

F =  1500 N

F/W = 500:147

Explanation:

Using the equation of motion, to get the initial velocity

v² = u²+2gs Equation 1

Where v =final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation

Given: v = 0 m/s ( at the maximum height), s = 2.5 m, g = -9.8 m/s²(against gravity)

substitute into equation 1

0² = u² +2×2.5×(-9.8)

u² = 49

u = 7 m/s.

a = u/t

Where t = time = 0.21 s

a = 7/0.21

a = 33.33 m/s²

Recall that,

F = ma Equation 2

Where F = force, m = mass of the impala.

Given: m = 45 kg and a =33.33 m/s²

Substitute into equation 2

F = 45(33.33)

F =1500 N.

Hence the force =1500 N.

Weight of the antelope = mg

W = mg

Where m = 45 kg, g = 9.8

W = 441 N.

F/W =1500/441

F/W = 500:147

Hence the ratio of force to antelope weight = 500:147

12000j is work done