Answer to Question #240380 in Physics for Morgan Hopkin

Question

Javier kicks a soccer ball 25 m/s at an angle of 44° from the top of a steep hill. The soccer ball lands at a point that is 12.9 m below the height from which it was kicked. Determine how long the ball was in the air and the horizontal displacement of the ball from Javier.

Answer 1

Find time upward:

$$t_u=\frac{v\sin\theta}{g}=1.77\text{ s}.$$

Time downward:

$$t_d=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\Big(\frac{(v\sin\theta)^2}{2g}+12.9\Big)}{g}}=2.4\text{ s},\\\space\\ T=t_u+t_d=4.17\text{ s}.$$

The horizontal displacement of the ball from Javier is

$$R=v_xT=v\cos\theta T=75\text{ m}.$$