COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 3, Problem 29QAP
To determine
The initial speed of the ball as it left Reggie's bat.
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You buy a plastic dart gun, and being a clever physics
student you decide to do a quick calculation to find its
maximum horizontal range. You shoot the gun straight
up, and it takes 5.6 s for the dart to land back at the
barrel.
• Part A
What is the maximum horizontal range of your gun?
Express your answer using two significant figures and include the appropriate units.
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Value
Units
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• From ta = -25 s until to = -15 s the plane is taxiing east along the runway at a constant speed of 4 m/s. From ts
until te = -10 s the plane slows down at a constant rate and comes to rest.
• From te until to =0 the plane turns around, turning nearly in place so that by to it is facing west.
• At to = 0 the plane is near the east end of the runway, at rest. From to until ti = 5.0 s it speeds up with a constant
acceleration going west. At t, it is going at a speed of 8.75 m/s.
• After t, the plane continues speeding up, but at a steadily reducing rate (i.e. it is still speeding up, but the magnitude
of its acceleration is decreasing). It lifts off at t2 = 65 s going at a speed of 85 m/s, having gone a distance of 2980
m down the runway.
• From t2 until t3 = 240 s the plane is climbing at an angle of 5.0° above horizontal, speeding up at a constant rate.
At t3 it is going at a speed of 230 m/s.
• At t3 the plane levels off, so it is flying at a constant height. From t3 until ta = 270 s…
QUESTION 6
Problem
For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height
hmax and the horizontal range R. b) For what value of 0 gives the highest maximum height?
Solution
The components of vo are expressed as follows:
Vinitial-x = vocos(e)
Vinitial-y = vosin(e)
а)
Let us first find the time it takes for the projectile to reach the maximum height.
Using:
Vfinal-y = Vinitial-y + ayt
since the y-axis velocity of the projectile at the maximum height is
Vfinal-y
Then
= Vinitial-y + ayt
Substituting the expression of vinitial-y and ay = -g, results to the following:
Thus, the time to reach the maximum height is
tmax-height =
We will use this time to the equation
Yfinal - Yinitial = Vinitial-yt + (1/2)ayt?
if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so
hmax = Vinitial-yt + (1/2)ayt2
substituting, the vinitial-y expression above,…
Chapter 3 Solutions
COLLEGE PHYSICS
Ch. 3 - Prob. 1QAPCh. 3 - Prob. 2QAPCh. 3 - Prob. 3QAPCh. 3 - Prob. 4QAPCh. 3 - Prob. 5QAPCh. 3 - Prob. 6QAPCh. 3 - Prob. 7QAPCh. 3 - Prob. 8QAPCh. 3 - Prob. 9QAPCh. 3 - Prob. 10QAP
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