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Chapter 2

Q. 2.8

Not a Bad Throw for a Rookie!

GOAL Apply the kinematic equations to a freely falling object with a nonzero initial velocity.

PROBLEM A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down, as shown in Figure 2.20. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, (d) the time needed for the ball to reach the ground, and (e) the velocity and position of the ball at t = 5.00 s. Neglect air drag.

STRATEGY The diagram in Figure 2.20 establishes a coordinate system with y_0 = 0 at the level at which the ball is released from the thrower’s hand, with y positive upward. Write the velocity and position kinematic equations for the ball, and substitute the given information. All the answers come from these two equations by using simple algebra or by just substituting the time. In part (a), for example, the ball comes to rest for an instant at its maximum height, so set v = 0 at this point and solve for time. Then substitute the time into the displacement equation, obtaining the maximum height.

F2.20

Step-by-Step

Verified Solution

(a) Find the time when the ball reaches its maximum height.

Write the velocity and position kinematic equations:

v  =  at  +  v_0
\Delta y  =  y  –  y_0  =  v_0t   +  \frac{1}{2} at^2

Substitute a  =  -9.80  m/s^2,  v_0  =  20.0  m/s,  and  y_0  =  0 into the preceding two equations:

(1) v  =  (-9.80  m/s^2)t  +  20.0  m/s
(2) y  =  (20.0  m/s)t  –  (4.90  m/s^2)t^2

Substitute v = 0, the velocity at maximum height, into Equation (1) and solve for time:

0  =  (29.80  m/s^2)t  +  20.0  m/s
t  =  \frac{-20.0  m/s}{-9.80  m/s^2}  =  2.04  s

(b) Determine the ball’s maximum height.

Substitute the time t = 2.04 s into Equation (2):

y_{max}  =  (20.0  m/s)(2.04  s)  –  (4.90  m/s^2)(2.04  s)^2  =  20.4  m

(c) Find the time the ball takes to return to its initial position, and find the velocity of the ball at that time.

Set y = 0 in Equation (2) and solve t:

0  =  (20.0  m/s)t  –  (4.90  m/s^2)t^2
=  t(20.0  m/s  –  4.90  m/s^2t)
t  =  4.08  s

Substitute the time into Equation (1) to get the velocity:

v  =  20.0  m/s  +  (-9.80  m/s^2)(4.08  s)  =  -20.0  m/s

(d) Find the time required for the ball to reach the ground.

In Equation (2), set y = -50.0 m:

-50.0  m  =  (20.0  m/s)t  –  (4.90  m/s^2)t^2

Apply the quadratic formula and take the positive root:

t  =  5.83  s

(e) Find the velocity and position of the ball at t = 5.00 s.

Substitute values into Equations (1) and (2):

v  =  (-9.80  m/s^2)(5.00  s)  +  20.0  m/s  =  -29.0  m/s

y  =  (20.0  m/s) (5.00  s)  –  (4.90  m/s^2)(5.00  s)^2  =  -22.5  m

REMARKS Notice how everything follows from the two kinematic equations. Once they are written down and the constants correctly identified as in Equations (1) and (2), the rest is relatively easy. If the ball were thrown downward, the initial velocity would have been negative.