Question

A startled armadillo leaps upward, rising 0.532 m in the first 0.202 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.532 m? (c) How much higher does it go? Use g=9.81 m/s2.

Answer 1

A) $d=v_0t+\frac{1}{2}at^2$
0.532 m = 0.202v_0 + ½(-9.81 m/s²)(0.202 s)²
v_0 = 3.62 m/s

B) $d = \frac{1}{2}(v+v_0)t$
0.532 m = ½(v + 3.62 m/s)(0.202) s
v = 1.65 m/s

C) $v^2=v_0^2+2ad$
At the highest point v = 0, so
0 = (1.65 m/s)^2 + 2(-9.81 m/s²)d
d = 0.139 m