Kinematic Equation Problem

Question

a. The maximum height above the ground of a projectile launched at an angle of 29.2° to the horizontal is found to be 57.5 m. What is the initial speed of the projectile?

b. What is the horizontal range of a projectile launched with a speed of 39.6 m/s at an angle of 28.2° above the horizontal?

c. A bicyclist is currently traveling south at 10 km/h and is located 17.3 km south of the center of the nearest town. She begins accelerating north at 4.01 km/h² and does so for 0.19 hours. What is the displacement of the bicyclist from the center of the nearest town at the end of the 0.19 hours of acceleration? Use the convention that north is positive, south is negative.

d. A runner is currently moving at a speed of 5.2 m/s. Over the time it takes the runner to travel the next 20.0 m she accelerates at a constant rate of 0.5 m/s². What is the runner's new speed at the end of the 20.0 m?

e. A projectile is launched horizontally with a speed of 22.1 m/s. If the projectile takes 4.85 seconds to reach the ground, how far above the ground was the launch point?

f. A ball is dropped from a building that is 28.1 m tall. Assuming the ball is in free fall, how many seconds will it take the ball to reach the ground?

g. A ball is thrown straight up with a speed of 15.1 m/s. Assuming free fall, how many seconds will it take to return to the original height at which it was released?

h. A ball is thrown straight downward with an initial speed of 6.1 m/s. How many meters below the release point will the ball be at a time of 4.2 seconds later if air resistance is ignored?

i. A boat travels 12.0 km east at a speed of 10 km/h, then travels 18.2 km north at a speed of 20 km/h and finally travels 12.3 km east at a speed of 15 km/h. What is the magnitude of the average velocity for the boat over the entire trip?

j. A boat travels 14.7 km east at a speed of 10 km/h, then travels 13.7 km north at a speed of 20 km/h and finally travels 14.4 km west at a speed of 15 km/h. What is the average speed for the car over the entire trip?

Answer 1

Hello Linda!

Wow, there's a lot of material here. I may have to return later to answer some of the other questions. Like any other physics problems, only a finite amount of skill can be acquired via reading and contemplating the material; Everything else is derived from practice.

a. First, we have a projectile launched at an angle of 29.2 degrees. We immediately know that there are x and y components of projectiles moving in an arc linear fashion. Movement in the x-direction is accomplished with a constant velocity. If we designate the upward y-direction as being positive, the component of motion in the y-direction is opposed by gravity when the projectile is moving upward. At some maximum height y, velocity = 0. Thus, two kinematic equations immediately come to mind: ( 1 ) y = v(initial )( t ) - ( 1 / 2 )( g )( t^2 ), and ( 2 ) v( final ) = v( initial ) - ( g )( t ). At this point, I'm unsure whether I will need to use both ( or other ) equations, but that's okay.

Since the projectile is launched at an angle of 29.2 degrees, either the sin ( theta ) component of velocity or the cos ( theta ) component of velocity must be used in either of the aforementioned equations. The component of velocity ( v ) that we are interested in moves in the verticle ( y ) direction. Therefore, ( v )( sin 29.2 degrees ) is the initial velocity component of interest to us.

Since distance ( d ) = ( v )( t ), where v = velocity, and t = time, we will need to know how much time it took for the projectile to reach a height of 57.5 m. If the final velocity of the projectile is zero at 57.5 m, from v ( final ) = v ( initial ) - ( g )( t ), we have 0 = ( v )( sin 29.2 ) - ( 9.8 m / s^2 )( t ). Therefore, -( v )( sin 29.2 ) = -( 9.8 m / s^2 )( t ). Solving for time gives us t = -( v sin 29.2 ) / -( 9.8 m / s^2 ).

Ooops! We were not given the value of v. So some other approach must be used :) .

Well, the other equation of interest states that y = ( v initial )( t ) - ( 1 / 2 )( g )( t^2 ). WHY NOT BEGIN ADDRESSING THIS PROBLEM AS A FREEFALL QUESTION? The projectile, after reaching a height of 57.5 m, will have a component of motion that will soon be in freefall in the y-direction. Therefore, our initial velocity at this point will be ZERO. Furthermore, we are now traveling in the ( -y ) direction.

-( 57.5 m ) = 0 - ( 1 / 2 )( g )( t^2 ). Now, -( 2 )( 57.5 m ) = -( g )( t^2 ). The negative terms cancel, and ( ( 115 m ) / ( 9.8 m / s^2 ) ) = t^2. Thus, the time ( t ) that the projectile falls in the y-direction is the square root ( sqrt ) of 11.7 m^2, and t = 3.43 s.

Recall that v( final ) = v( inital ) - ( g )( t ). We now have a value for t, so we can solve for v( final ). By symmetry, our value of the projectile's velocity will be the NEGATIVE of what it was when it began its trip upward. Therefore, - v( final ) = 0 - ( 9.8 m / s^2 )( 3.43 s ). Our final velocity of -( v ) = - 33.6 m / s.

If 33.6 m / s was our initial velocity in the y-direction, then 33.6 m / s = ( v )( sin 29.2 ). Solving for the inital velocity ( v ) gives us v = ( ( 33.6 m / s ) / (sin 29.2 ) ) = 68.9 m / s.

CHECK: If y = ( v initial )( t ) - ( 1 / 2 )( g )( t^2 ), then 57.5 m = ( v sin theta )( t ) - ( 1 /2 )( g )( t^2 ). Substitution gives us 57.5 m = ( 68.9 m / s )( sin 29.2 )( 3.43 s ) - ( 1 / 2 )( 9.8 m / s^2 )( 3.43 s )^2.

y = ( 115.3 m ) - ( 57.6 m ) = 57.7 m.

WELL, I'M DONE FOR THIS EVENING. Enjoy your studies!