Question

Leaping Armadillo A startled armadillo leaps upward, rising 0.570 m in the first 0.200 seconds.

(a) What is its initial speed as it leaves the ground?
  m/s
(b) What is its speed at the height of 0.570 m?
m/s
(c) How much higher does it go?
m

Answer 1

(a) If the initial velocity v_{​​​​​​i} then at time t expression of height y is given by ,

So at y= 0.570m ; t= 0.200sec a

And let g= - 9.8m/s^{​​​​2}​​​​​ (as gravitation acts in downward direction)

So initiail speed as it leaves the ground is ,

(b) expression of speed at height y is given by,

Speed at height 0.570m is

(c) at maximum height its its velocity becomes zero. So

​​​​​

So it goes 0.748m higher.