Answer to Question #135079 in Physics for Shakera

Question

A football is kicked with an initial velocity of 29 m/s at an angle of 45 degrees with the horizontal. Determine the time of flight,horizontal distance,and the peak height of the football

Answer 1

Given:

$v_0=29\:\rm m/s$

$\theta=45^{\circ}$

(a) the time of flight

$$t=\frac{2v_0\sin\theta}{g}=\frac{2\times29\times \sqrt{2}/2}{9.8}=4.2\:\rm s$$

(b) the horizontal distance

$$R=\frac{v_0^2\sin2\theta}{g}=\frac{29^2\times 1}{9.8}=85.8\:\rm m$$

(c) the peak height

$$H=\frac{v_0^2\sin^2\theta}{2g}=\frac{29^2\times 1/2}{2\times9.8}=21.5\:\rm m$$