Answer to Question #91268 in Mechanics | Relativity for Umar

Question

In 1978, Geoff Capes of the United Kingdom won a competition for throwing 5 Ib bricks; he threw one brick a distance of 44.0m. Suppose the brick left Capes hand at an angle of 45.0 degree with respect to the horizontal.
a) what was the initial speed for the brick? (20.8 m/s)
b) what was the maximum height reached by the brick? (11.0m)

Answer 1

The range of a projectile

$$R=\frac{v_i^2\sin 2\theta}{g}$$

(a) So, the initial speed

$$v_i=\sqrt{gR/\sin 2\theta}=\sqrt{9.8\times 44.0/\sin 90^{\circ}}=20.8\:\rm{m/s}$$

(b) The maximum height

$$h=\frac{R\tan \theta}{4}=\frac{44.0\times\tan 45^{\circ}}{4}=11.0\:\rm{m}$$