Question #271474

A volleyball player hits the ball 40 cm above the top of the net towards his opponents’ court at a velocity of 15.0 m/s along the horizontal. If the top of the net is 2.0 m high from the floor and the length of the opponents’ court is 9.0 m, will the volleyball land within the opponents’ court? Why?




1
Expert's answer
2021-11-25T16:23:28-0500

h0=2mh_0 = 2m

h1=40cm=0.4mh_1= 40cm =0.4m

s1=9ms_1 = 9m

v0=15msv_0 = 15\frac{m}{s}

g=9.8ms2g = 9.8\frac{m}{s^2}

h=h0+h1=2.4mh = h_0+h_1 = 2.4m

Ball movement time:\text{Ball movement time:}

t=2hg=22.49.80.69st = \sqrt\frac{2h}{g}=\sqrt\frac{2*2.4}{9.8}\approx0.69s

s=v0t=150.69=10.35ms = v_0t= 15*0.69= 10.35m

s>s1s>s_1

The distance traveled by the ball before falling is\text{The distance traveled by the ball before falling is}

greater than the length of the court\text{greater than the length of the court}


Answer: The ball will not hit the court\text{Answer: The ball will not hit the court}


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