Question

A startled armadillo leaps upward, rising 0.540 m in the first 0.216 s. (a) What is its initial speed as it leaves the ground? (b)What is its speed at the height of 0.540 m? (c) How much higher does it go? Use g=9.81 m/s^2.

Answer 1

a) 3.6 m/s

b) 1.53 m/s

c) 0.12 m

Step-by-step explanation:

t = Time taken = 0.216 s

u = Initial velocity

v = Final velocity

s = Displacement = 0.54 m

a = Acceleration due to gravity = 9.81 m/s² (negative upward)

`s=ut+(1)/(2)at^2\\\Rightarrow 0.54=u* 0.216+(1)/(2)* -9.81* 0.216^2\\\Rightarrow u=(0.54+(1)/(2)* 9.81* 0.216^2)/(0.216)\\\Rightarrow u=3.6\ m/s`

Initial speed as it leaves the ground is 3.6 m/s

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* -9.81* 0.54+3.6^2)\\\Rightarrow v=1.53\ m/s`

Speed at the height of 0.540 m is 1.53 m/s

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-3.6^2)/(2* -9.81)\\\Rightarrow s=0.66\ m`

The total height the armadillo leaps is 0.66 m

So, the additional height is 0.66-0.54 = 0.12 m