Question

An athlete executing a long jump leaves the ground at a 28.0 angle and travels 7.80 m. (a) What was the takeoff speed? (b) If this speed were increased just 50%, how much longer would the jump be?

Answer 1

Given:

$\theta=28.0^{\circ}$

$R=7.80\:\rm m$

(a) The range

$$R=\frac{v_0^2\sin2\theta}{g}$$

So, the initial speed

$$v_0=\sqrt{\frac{gR}{\sin2\theta}}=\sqrt{\frac{9.8*7.8}{\sin2*28^{\circ}}}=9.6\:\rm m/s$$

(b) The new range

$$R_1=\frac{(1.5v_0)^2\sin2\theta}{g}=1.5^2R=1.5^2*7.8=17.6\:\rm m$$