A child in a car tosses a ball upward so that after release it requires 1/2 second to rise and fall back into the child's hand at the same height from which it was released.  The car is traveling at a constant speed of 10 meters / second in the horizontal direction. 

• Between release and catch, how far did the ball travel in the horizontal direction? 

The ball continues to travel at 10 m/s in the horizontal direction and in 1/2 second moves `ds = vAve * `dt = 10 m/s * .5 s = 5 m in that direction.

As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught? 

Relative to the roadside the ball travels at a constant horizontal velocity while rising more and more slowly, then falling more and more quickly. This gives it a curved path within that frame of reference. In fact it can (and will) be shown that the path would be parabolic, for the present a good description would indicate a curved path, rising or increasing at a decreasing rate then falling or decreasing at an increasing rate.

• Sketch the path of the ball as observed by a line of people standing along the side of the road.  Describe your sketch.  What was shape of the path of the ball?

The ball travels along a parabolic path.

• How fast was the ball moving in the vertical direction at the instant of release?  At that instant, what is its velocity as observed by a line of people standing along the side of the road?

In the vertical direction the ball has displacement `ds = 0 and accelerates downward at 9.8 m/s^2 for .5 s.  Letting the upward direction be positive we have `ds = 0, a = -9.8 m/s^2 and `dt = .5 s.

• Reasoning directly:  In .5 s its change in velocity is -9.8 m/s^2 * .5 s = -4.9 m/s.  Its displacement is zero so its average velocity is 0; thus its initial and final velocities are equal and opposite.  The equal and opposite quantities which change by -4.9 m/s are 2.45 m/s and -2.45 m/s.  So the initial vertical velocity is 2.45 m/s.

• The same results can be obtained using the equations of unif accel motion, with a = -9.8 m/s^2, `ds = 0 and `dt = .5 s.  The third equation of motion yields v0 = (`ds - .5 a `dt^2) / `dt = ( 0 - .5 * (-9.8 m/s^2) * (.5 s)^2) / (.5 s) = 2.45 m/s.

The velocity of the ball at this instant has horizontal component 10 m/s and vertical component 2.45 m/s.

• The magnitude of the velocity is therefore sqrt( (10 m/s)^2 + (2.45 m/s)^2) = 10.3 m/s, approx.

• The velocity is directed at angle arcTan(2.45 m/s / (10 m/s) ) + arcTan(.245) = 14 deg, approx., with respect to the horizontal direction.

• How high did the ball rise above its point of release before it began to fall back down?

Using v0 = 2.45 m/s, vf = 0 and a = -9.8 m/s^2: 

• Direct reasoning:  Average velocity is (2.45 m/s + 0) / 2 = 1.23 m/s.  Change in velocity is -2.45 m/s and the time required for the change is `dt = `dv / a = -2.45 m/s / (9.8 m/s^2) = .25 s.  Displacement is vAve * `dt = 1.23 m/s * .25 s = .31 m.

Some student comments:

STUDENT COMMENT I found it interesting how the horizontal and vertical velocities were added together and the angle found.. Because technically velocity is speed in a given direction correct?

INSTRUCTOR RESPONSE

Your statement is correct.

Velocity is a vector quantity, having a magnitude and a direction. On a straight line, the direction is merely positive or negative. In two dimensions the direction is specified by the direction of the vector, which at this point in our course is specified by the angle of the vector as measured counterclockwise from the positive x axis.

The magnitude of velocity is speed, so it is accurate to say that velocity is speed in a given direction.