Question #240380

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.


1
Expert's answer
2021-09-23T08:32:43-0400

Find time upward:


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

Time downward:


td=2hg=2((vsinθ)22g+12.9)g=2.4 s, T=tu+td=4.17 s.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=vxT=vcosθT=75 m.R=v_xT=v\cos\theta T=75\text{ m}.


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