Answer to Question #94157 in Mechanics | Relativity for Saud

Question

When startled, an armadillo will leap upward. Suppose it
rises 0.544 m in the first 0.200 s. (a) What is its initial speed as it
leaves the ground? (b) What is its speed at the height of 0.544 m?(c) How much higher does it go?

Answer 1

$T = 0.200 s; H = 0.544 m; \\ v_0 = ?; V = ?; h_{max} = ? \\ h = v_0t - \frac{1}{2}gt^2; \\ v_0 = \frac{2H + gT^2}{2T} = \frac{2\cdot0.544m + 9.81\frac{m}{s^2}(0.200s)^2}{2\cdot0.200s} = 3.701 \frac{m}{s}; \\ v = v_0 - gt ; \\ V = v_0 - gT = \frac{2H - gT^2}{2T} = \frac{2\cdot0.544m - 9.81\frac{m}{s^2}(0.200s)^2}{2\cdot0.200s} = 1.739 \frac{m}{s}; \\ h_{max} = \frac{v_0^2}{2g} = \frac{(3.701 \frac{m}{s})^2}{2*9.81\frac{m}{s^2}} \approx 0.698 m.$

Answers:

(a) 3.701 m/s

(b) 1.739 m/s

(c) approx 0.698 m (i.e. from the ground).