Q: A soccer player kicks the ball toward a goalie that is 29.0 m in front of him. The ball leaves his foot at a speed of 19.5 m/s and at an angle of 33.4° relative to the ground. Find the speed of the ball when the goalie catches it in front of the net. ( Note: The answer is not 19.5 m/s, because the ball doesn’t fully return to the ground. )
A: When the ball is caught by the goalie, it descends from the sky diagonally, and its final horizontal and vertical velocity components will be ( + x ) and ( – y ). This tells us that the initial velocity was in a diagonal direction with ( + x ) and ( + y ) components horizontal and vertical components. Since the angle between the initial velocity vector ( 19.5 m/s ) and the ground is 33.4°, we can calculate its horizontal ( Vx ) and vertical ( Vy ) components.
The value of Vy = V sin 33.4°, and Vx = V cos 33.4°. Additionally, we must know how much time ( t ) it takes for the ball to reach the goalie. By using this derived value of time, we can calculate the final velocity of the ball at the position where it is caught ( as opposed to when it hits the ground, which would simply be the exact opposite of its value upon leaving the ground ).
Recall that the soccer player is 29.0 m away from the goalie. Since ( d ) = ( v )( t ), the time that will pass as the ball travels to the goalie is t = [ ( 29.0 m ) / ( Vx ) ]. Solving for ( t ), we have ( t ) = [ ( 29.0 m ) / ( V cos 33.4° ) ] = [ ( 29.0 m ) / ( 16.3 m / s ) ] = 1.78 s.
IT IS ALMOST ALWAYS GOOD TO FIRST DETERMINE HOW TIME IS RELATED TO KINEMATIC PROCESSES.
The initial velocity component in the ( y ) direction is ( Vy initial ) = ( V sin 33.4° ) = [ ( 19.5 m / s )( sin 33.4° ) ] = 10.7 m/s.
We now use ( Vy ) and the total time of flight to determine the ( y ) component of velocity experienced where the goalie is positioned: The appropriate equation is ( Vy final ) = ( Vy initial ) – ( g )( t ). Therefore, ( Vy final ) = ( 10.7 m / s ) – ( 9.8 m / s^2 )( 1.78 s ) = – 6.74 m / s.
WHEN THE TIME OF FLIGHT IS ½ OF THE TRIP, Vy FINAL = 0, AND THE BALL IS AT ITS APEX. SINCE WE DETERMINED THE TOTAL TIME OF FLIGHT FROM THE BEGINNING, Vy FINAL = Vy INITIAL – gt GIVES US THE NEGATIVE Y-COMPONENT WE’RE LOOKING FOR.
We now have values of ( Vy final ) and ( Vx ). We use the Pythagorean theorem of [ x^2 + y^2 = z^2 ] to determine the value of V: If ( V final )^2 = ( Vx )^2 + ( Vy )^2, then ( V final ) = √ ( Vx^2 + Vy^2 ).
V final = √ [ ( 16.3 m / s )^2 + ( – 6.74 m / s )^2 ] = 17.6 m / s.
FOLLOW-UP QUESTION: How far in the ( x ) direction would the ball have traveled if the goalie hadn’t caught it?
A: When a projectile traveling in curvilinear motion reaches its apex, it will have ceased its upward motion against gravity. For a moment, ( Vy final ) = 0. Additionally, the projectile will have traveled half the distance it will travel in the ( x ) direction. If we use the equation ( Vy final ) = ( Vy initial ) – ( g )( t ), we can solve for the time needed to complete half the trip.
If ( Vy final ) = 0 when the ball has traveled half the distance ( x ), then 0 = ( Vy initial ) – ( g )( t ). Subtracting ( Vy initial ) from both sides, we have -( Vy initial ) = -( g )( t ). The negative signs cancel, and we divide both sides by the gravitational constant of acceleration ( g ) to solve for ( t ). Thus, ( t ) = ( Vy initial / g ) = [ ( 10.7 m / s ) / ( 9.8 m / s^2 ) ] = 1.09 s. Therefore, the total time ( t ) of flight = 2.18 s. The total distance traveled in the ( x ) direction is ( x ) = ( Vx )( t total ) = ( 16.3 m / s )( 2.18 s ) = 35.6 m.
RENAISSANCE PHYSICS 