An impala is an african antelope capable of a remarkable vertical leap in one recorded leap, a 45kg impala went into a deep crouch, pushed straight up for .21s. and reached a height of 2.5 m above the ground. To achieve this leap what force did the impala push down on the ground. What is the ratio of the force with the impalas weight
As the impala rises to its maximum height, its velocity will decrease to 0 m/s at the rate of 9.8 m/s each second. Let’s use the following equation to determine the impala’s after the force is exerted.
vf^2 = vi^2 + 2 a d, vf = 0, a = -9.8, d = 2.5
0 = vi^2 + 2 -9.8 2.5
vi^2 = 49
v = 7 m/s
The force must increase the impala’s velocity from 0 m/s to 7 m/s in 0.21 second
F = m * a
a = 7 ÷ 0.21 = 33⅓ m/s^2
F = 45 * 33⅓ = 1500 N
Let’s determine the impala’s weight.
Weight = 45 * 9.8 = 441 N
Ratio = 1500: 441
This is approximately 3.4 to 1.
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Impulse vector = Î = F̂ₐᵥₑΔt = Δmv̂ = mΔv̂
the deer's velocity in time Δt changes from 0 to vₒ. vₒ is the initial velocity for it's vertical jump
for vₒ we have 2.5 m = maximum height = vₒ²/[2g]
=> vₒ = √[(2)2.5(9.81)] = 7 m/s (upwards)
Fₐᵥₑ(0.21) = mΔv = 45[7 - 0] = 315.16
=> Fₐᵥₑ = 315.16 / .21 = 1500.77 N (downwards opp. the direction of vₒ)
Fₐᵥₑ / mg = 1500.77 / [45x9.81] ~= 3.4/1
[please note: derivation of formula H = maximum height = vₒ²/[2g]
where vₒ is the initial vertical velocity. at the high point of a projectile the vertical velocity is = 0 as it reverses direction groundwards.
=> from v² = u² + 2as
=> 0² = vₒ² -2gH or H = vₒ²/[2g] ]
hope this helps