The horizontal distance x traveled by a soccer ball if it is kicked from ground level with an initial velocity of v and angle θ is x= (1/32)v^2*sin 2θ. Find the distance traveled if the initial velocity is 50 feet per second and the angle is 30°?

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Answer 1 · 2018-03-19T07:27:30

The distance is #=67.66ft#

The equation for the horizontal distance is

#x=1/32v^2sin(2theta)#

The initial velocity is #v=50fts^-1#

The angle is #theta=30^@#

#sin60^@=sqrt3/2=0.866#

Therefore,

#x=1/32*50^2*sin(2*30^@)#

#x=1/32*50^2*sin(60^@)#

#=1/32*2500*0.866#

#=67.66ft#