An impala is an African antelope capable of a remarkable vertical leap. In one recorded leap, a 25.5 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?

The impala pushed down the ground with a force of 37.49N

Explanation:

The force at which the impala pushed down the ground can be calculated for using the Newton's second law of motion,

Force = mass × acceleration

Force = mass ×(velocity/time)

Given mass = 25.5kg

Time = 0.21seconds

To get the velocity, we will use one of the equation of motions;

Using v² = u²+2gH

where;

H is the height reached from the ground = 2.5m

g = 9.81m/s²

u is the initial velocity = 0m/s

v is the final velocity=?

Substituting this values to get the final velocity v;

v² = 0²+2(9.81)(2.5)

v² = 49.05

v = √49.05

v = 7.0m/s

Substituting this velocity into the formula for force we have;

Force = 25.5×(7.0/0.21)

Force = 25.5 × 1.47

Force = 37.49N

The impala pushed down the ground with a force of 37.49N

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