Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

impala need 1099N force to push down on the ground

Explanation:

The final height is 2.5m and

final velocity is 0

Calculate the initial velocity

mgh = 1/2mv²

v² = 2gh

   = 2(9.8)(2.5)

   = 7m/s

The initial velocity is gain in 0.21s

7m/s =a × 0.21s

a = 7 / 0.21

a = 33.3m/s

Using the following formula to calculate the force

$F_{net} - mg = ma\\\\F_{net}= 25.5\times33.3+25.5\times9.8\\\\F_{net} = 1099N$

Thus,  impala need 1099N force to push down on the ground