Question

In a soccer practice session the football is angle is kept at the centre of the field 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of 64 ft/s at an angle of ${45}^2$ to the horizontal. Will the ball reach the goal post ?

Answer 1

Given, g = 9.8 $m/s^2=32.2ft/s^2$

40 yd = 120 ft

Horizontal range,

x = 120 ft, u = 64 fr/s,

$\theta ={45}^0$

We know that horizontal range,

x = u cos~ $\theta t$

$\Rightarrow t=\frac{x}{ucos\theta }$

= $\frac{120}{64cos{45}^0}=2.65sec$

y = $sin\theta \left(t\right)-\frac{1}{2}\left)\right(32.2\left)\right(2.65{)}^{2}sec$

= 64 $\left(\frac{1}{\sqrt{2}}\right)\left(2.65\right)-\frac{1}{2}\left(32.2\right)(2.65{)}^2$

= 7.08 ft which is less than the 8th of goal post

In time 2.65 s the ball trasvels horizontal distance 120 ft (40 yd) and verticle height 7.08 ft which is less than 10 ft. The ball will reach the goal post