An impala is an african antelope capable of a remarkable vertical leap. 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? express your answer to two significant figures and include the appropriate units. fnet = what is the ratio of this force to the antelope's weight? express your answer using two significant figures. fnet / w =

F = 51 10² N and The force applied is 11.6 times greater than the weight

Explanation:

As we are asked the force we can use Newton's second law and we calculate the acceleration by kinematics

      y = v₀ t + ½ a t²

On the ground its initial velocity is zero (v₀ = 0)

      y = ½ a t²

      a = 2y / t²

      a = 2 2.5 / 0.21²

      a = 113.38 m / s²

We calculate the force

      F = ma

      F = 45  113.38

      F = 5102 N

      F = 51 10² N

For the relationship of this force with the weight, divide the two

      F / W = 51 10²/45 9.8

      F / W = 11.6

The force applied is 11.6 times greater than the weight

Explanation:

As we know that impala is pushed upwards straight up with uniform acceleration

So here we can use kinematics equation

So we will have

now we have

Now ratio of this force with the weight is given as

trig 15.3 cos