Question

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 =

Answer 1

`F = 5100 N`

`R = 11.56`

Step-by-step 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

`y = vt + (1)/(2)at^2`

`2.5 = (1)/(2)a(0.21)^2`

`a = 113.4 m/s^2`

now we have

`F = ma`

`F = 45(113.4)`

`F = 5100 N`

Now ratio of this force with the weight is given as

`R = (F)/(F_g)`

`R = (5100)/(45 * 9.81)`

`R = 11.56`

Answer 2

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

Step-by-step 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