An impala is an african antelope capable of a remarkable vertical leap

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

4 Answers

  • 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

  • Crouch Height

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