Question #184680

During a cricket match a cricket ball is hit with an initial velocity of 45.0 m s-1 at an angle of 30.0o to the horizontal from a height of 1.10 m above the ground. It lands in the spectators' balcony which is 8.50 m above the ground. 

Calculate the horizontal and vertical components of the cricket ball’s initial velocity.


1
Expert's answer
2021-04-26T17:09:33-0400

Answer

Initial horizontal velocity

vx=45.0(cos30°)=38.97m/sv_x=45.0(cos30°)=38.97m/s


Initial vertical velocity

vx=45.0(sin30°)=22.5m/sv_x=45.0(sin30°) =22.5m/s





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