Question #17331

Ball is kicked at 20 degree angle of a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.0 seconds later, what was the initial speed given to the ball? Assume the speed of sound in air to be a constant 343 m/s

Expert's answer

v0sina=Vy

vy=gt1

t1=2Vy/g

gt2^2/2=h

t2=Sqrt(2h/g)

v0cosat=S=Vs*t3

t3=(v0/vs)cosa(t1+t2)

t1+t2+t3=3s=T

((V0/vS)cosa+1)(2V0sina/g+sqrt(2h/g))=T

v0^2(sin2a/gVs+(2sina/g+sqrt(2h/g)*cosa/vs)V0+sqrt(2h/g)-T

v0=2.24

vy=gt1

t1=2Vy/g

gt2^2/2=h

t2=Sqrt(2h/g)

v0cosat=S=Vs*t3

t3=(v0/vs)cosa(t1+t2)

t1+t2+t3=3s=T

((V0/vS)cosa+1)(2V0sina/g+sqrt(2h/g))=T

v0^2(sin2a/gVs+(2sina/g+sqrt(2h/g)*cosa/vs)V0+sqrt(2h/g)-T

v0=2.24

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