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

Answer 1

Impulse vector = F̂ₐᵥₑΔt = Δmv̂ = mΔv̂

[the deer pushes the ground with an average force Fₐᵥₑ for time = Δt s and then gains a vertical velocity v from rest as the ground exerts the same force on the deer upwards as per Newton's III law ]

the deer's velocity in time Δt changes from 0 to v

Fₐᵥₑ(0.21) = Δmv̂ = mΔv = 45[v - 0] = 45v --------------------> (i)

also from Energy Conservation => the kinetic energy of the deer is changed to potential energy at the highest point of the leap.

=> k.e = ½mv² = p.e = mgh

=> v² = 2gh => v = √(2gh) = √(2(9.81)(2.5)) = √49.05 = 7.004 m/s

the impala creates an acceleration in the .21 seconds it is leaving the ground... (7-0)/.21=33.33m/s^2

new acceleration=9.8+33.3=43.1m/s^2

F=45kg*43.1m/s^2=1939.5N