GOAL Apply the kinematic equations to a freely falling object with a nonzero ini
ID: 1327812 • Letter: G
Question
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 the figure. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) thetime 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 the figure establishes a coordinate system with y0 = 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.
SOLUTION
(A) Find the time when the ball reaches its maximum height.
Write the velocity and position kinematic equations.
Substitute a = 9.80 m/s2, v0 = 20.0 m/s, and y0 = 0 into the preceding two equations.
(1)
v =
9.80 m/s2
t + 20.0 m/s
(2)
y = (20.0 m/s)t
4.90 m/s2
t2
Substitute v = 0, the velocity at maximum height, into Equation (1) and solve for time.
(B) Determine the ball's maximum height.
Substitute the time t = 2.04 s into Equation (2).
(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.
Substitute the time into Equation (1) to get the velocity.
v = 20.0 m/s +
9.80 m/s2
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/s2
t2
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).
LEARN MORE
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.
QUESTION How would the answer to part (b), the maximum height, change if the person throwing the ball jumped upward at the instant he released the ball?
The maximum height would increase.The maximum height would remain the same. The maximum height would decrease.
PRACTICE IT
Use the worked example above to help you solve this problem. A ball is thrown from the top of a building with an initial velocity of 20.8 m/s straight upward, at an initial height of 55.9 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure.(a) Determine the time needed for the ball to reach its maximum height.
s
(b) Determine the maximum height.
m
(c) Determine 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) Determine the time needed for the ball to reach the ground.
s
(e) Determine the velocity and position of the ball at t = 5.01 s.
EXERCISEHINTS: GETTING STARTED | I'M STUCK!
A projectile is launched straight up at 70 m/s from a height of 80 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.
(a) Find the maximum height of the projectile above the point of firing.
m
(b) Find the time it takes to hit the ground at the base of the cliff.
s
(c) Find its velocity at impact.
m/s
Explanation / Answer
the maximum height of the projectile above the point of firing is
h' = h + 80 m
h = u^2/ 2g = ( 70 m/s)^2/ 2( 9.8)
= 250 m + 80 m = 330 m
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(b)
the time it takes to hit the ground at the base of the cliff is
t = root 2h'/g = 2 ( 330 m)/ ( 9.8) = 8.2 s
the time is
T = t+ t' = 8.2 s + v/g = 8.2 s + ( 70 m/s)/9.8 = 15.34 s
(c)
from the kinmatic equation
v = gT = (9.8) ( 15.34 s) = 150.36 m/s
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