Question #287040

 A golf ball is launched from the roof of a school with a velocity of 20 m/s at an angle of 300 above the horizontal. If the roof is 40 m above the ground, calculate:

a) the ball’s time of flight

b) the ball’s horizontal displacement




1
Expert's answer
2022-01-14T09:48:57-0500

a) The time of flight:


t=vsinθg+2(h+((vsinθ)2/(2g))g=4.05 s.t=\dfrac{v\sin\theta}{g}+\sqrt{\dfrac{2(h+((v\sin\theta)^2/(2g))}{g}}=4.05\text{ s}.


b) The range:


R=vcosθt=70.2 m.R=v\cos\theta·t=70.2\text{ m}.


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