Question

An impala is an African antelope capable of a remarkable vertical leaf. 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? What is the ratio of this force to the antelope's weight?

Answer 1

The force required to push the impala down on ground is 1,500 N.The ratio of force to weight is 3.4.

Answer:

Explanation:

The mass of the impala is given as 45 kg, the height or displacement of the impala is 2.5 m and it took 0.21 s. As the impala was moved straight up, so the acceleration due to gravity will be working as acceleration but in opposite direction.

$2as=v^{2}- u^{2}$

As the initial velocity u is given as 0 and the final velocity we have to find

$0-v^{2}=-2*9.8*2.5$

$49=v^{2}$

So v = 7 m/s.

As the final velocity is 7 m/s and the time is 0.21 s, then acceleration is rate of change of velocity with time.

$a = \frac{v}{t} =\frac{7}{0.21}=33.333 m/s^{2}$

As per Newton's second law of motion, the force acting on the impala will be equal to the product of mass and acceleration of the impala

F = Mass * Acceleration = 45 * 33.333=1,500 N.

So the force required to push the impala down on ground is 1,500 N.

And the weight is measured as product of acceleration due to gravity with mass.

Weight = 45 * 9.8 = 441.

So the ratio of force to weight is 1500/441 = 3.4

Thus, the ratio of force to weight is 3.4.