A startled armadill leaps upward rising .536 m in the first .197 s. What is its initial speed as it leaves the ground? What is its speed at the height f .536m? How much higher does it go?
y(t) = V*t - (g/2)t^2
where V is the initial velocity leaving the ground.
At t = 0.197 s
0.536 = V*(0.197)- 4.9*(0.197)^2
Solve that equation for V
The relationship between instantaneous velocity v and height y is given by the enery equation
M g y + (1/2) M v^2 = (1/2) M V^2
The M's cancel out.
v^2 = V^2 - 2 g y
For the maimum height ymax, subsitute 0 vor v and solve for y in the above equation.