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. To achieve this vertical leap, with a. 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?

(a) F = 1500 N.

(b) Ratio force to the antelepe's weight = 3.40

Explanation:

Force : This can be defined as the product of mass and the distance moved by a body. Its S. I unit is Newton. It can be represented mathematically as

F = Ma

Where F = force, M = mass (Kg) and a = Acceleration (m/s²)

Weight: This can be defined as the force on a body due to gravitation field. It is also measured in Newton (N). It can be represented mathematically as

W = Mg

Where W = weight of the body, M = mass of the body (Kg), g = Acceleration due to gravity.

(a)

F = Ma

Where M = 45kg,

a = unknown.

But we can look for acceleration Using one of the equation of motion,

v² = u² + 2gs

Where v = final velocity (m/s), u = initial velocity (m/s) g = 0 m/s, g = 9.8m/s² and s = height = 2.5m.

∴ v² = 2gs

v = √2gs = √ (2*9.8*2.5)

v = √49 = 7m/s

With the force applied, the impala's velocity must increase from 0 m/s to 7 m/s in 0.21 second

∴ a = (v-u) / t

a = (7-0) / 0.21 = 7/0.21

a = 33.33 m/s².

F = 45 * 33.33 ≈ 1500

F = 1500 N.

(b)

Where F = Force = 1500 N

and W = Weight = Mg = 45 * 9.8 = 441 N

∴Ratio force to the antelepe's weight = F/W = 1500/441 = 3.40

Ratio force to the antelepe's weight = 3.40