Answer to Question #184680 in Field Theory for Sebastien

Question

During a cricket match a cricket ball is hit with an initial velocity of 45.0 m s^-1 at an angle of 30.0^o 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.

Answer 1

Answer

Initial horizontal velocity

$v_x=45.0(cos30°)=38.97m/s$

Initial vertical velocity

$v_x=45.0(sin30°) =22.5m/s$