Q: The highest barrier that a projectile can clear is 13.7 m, when the projectile is launched at an angle of 10.5° above the horizontal. What is the projectile’s launch speed?
A: Recall that a projectile that travels in a curvilinear path has components of motion in the ( x ) and ( y ) direction. At our convenience, we can analyze these components of motion selectively to obtain information that is pertinent to solving many problems. The information that is immediately noteworthy pertains to the fact that the maximum height that an object travelling in curvilinear motion will reach is located at the apex of the curved path. At this location, 1/2 of the total distance travelled in the ( x ) direction has been travelled.
When an object travels parabolically against the force of gravity, its vertical ( y ) component of velocity ( Vy final ) will be zero when it reaches its apex. Since we know the height ( y ) of the barrier being surpassed, and we have an equation that relates ( Vy final ), ( Vy initial ), and ( y ), we can use it to solve for ( Vy initial ).
Our equation of choice is ( Vy final )^2 = ( Vy initial )^2 + ( 2 )( a )( y ). Take note that the acceleration due to gravity opposes the upward motion towards the apex. Since the vertical ( y ) direction has been assigned a positive value, our value for the acceleration ( a ) due to gravity is ( a ) = ( – g ). Substituting ( – g ) into the prior equation gives us ( Vy final )^2 = ( Vy initial )^2 – ( 2 )( g )( x ).
Since the final velocity in the vertical direction ( Vy final ) is equal to zero, we can solve for ( Vy initial ) as follows: 0 = ( Vy initial )^2 – ( 2 )( g )( x ). This leads to -( Vy initial )^2 = -( 2 )( g )( x ). The negative signs cancel, and ( Vy initial )^2 = ( 2 )( 9.8 m / s^2 )( 13.7 m ).
( Vy initial ) = sqrt [ ( 2 )( 9.8 m / s^2 )( 13.7 m ) ] = 16.4 m / s.
We can now have a horizontal vector that is opposite of our angle of interest ( theta ). Since the original value of velocity ( V ) is the hypotenuse of our system, we can use the sin ( theta ) function to solve the problem.
Sin ( theta ) = ( opposite / hypotenuse ) = ( Vy initial / V ). Solving for ( V ), we have ( V ) = [ ( Vy initial ) / sin ( theta ) ]. So ( V ) = [ ( 16.4 m / s ) / ( sin 10.5 ) ] = 90.0 m / s.
Check: Sin ( theta ) = ( opp / hype ) = [ ( 16.4 m / s ) / ( 90 m / s ) ]. Thus, ( theta ) = sin^-1 [ ( 16.4 m / s ) / ( 90 m / s ) ] = 10.5 degrees.
RENAISSANCE PHYSICS 