Problem#1
When baseball players throw the ball in from the outfield, they usually allow it to take one bounce before it reaches the infield, on the theory that the ball arrives sooner that way. Suppose that the angle at which a bounced ball leaves the ground is the same as the angle at which the outfielder threw it, as in Figure 1, but that the ball’s speed after the bounce is one half of what it was before the bounce.
(a) Assuming the ball is always thrown with the same initial speed, at what angle ' should the fielder throw the ball to make it go the same distance D with one bounce (blue path) as a ball thrown upward at 45.0° with no bounce (green path)? (b) Determine the ratio of the times for the one-bounce and no-bounce throws.
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| Fig.1 |
Answer:
The special conditions allowing use of the horizontal range equation applies.
R = vi2 sin 2θ/g
For the ball thrown at 45°,
X45 = D = vi2 sin 900/g = vi2/g
For the bouncing ball,
D = R1 + R2
D = [vi2 sin2θ/g] + [(vi/2) 2 sin 2θ/g]
where θ is the angle it makes with the ground when thrown and when bouncing.
(a) We require:
vi2/g = [vi2 sin2θ/g] + [(vi/2) 2 sin 2θ/g]
1 = sin 2θ + sin 2θ/4 = 5 sin 2θ/4
Sin 2θ = 4/5
2θ = sin-1(4/5) = 53.2
θ = 26.60
(b) The time for any symmetric parabolic flight is given by
yf = yi + vyit + ½ ayt2
0 = 0 + vi sin θit – ½ gt2
0 = t(vi sin θi – ½ gt)
t = 0 is the time the ball is thrown and t = 2vi sin θi/g is the time at landing.
So for the ball thrown at 45.0°,
t45 = 2vi sin 450/g = 1.414vi/g
For the bouncing ball,
t = t1 + t2
t = 2vi sin 26.60/g + 2(vi/2) sin 26.60/g = 3vi sin 26.60/g
t = 1.343vi/g
The ratio of this time to that for no bounce is
t/t45 = (1.343vi/g)(g/1.414vi)
t/t45 = 0.949
Problem#2
A boy can throw a ball a maximum horizontal distance of R on a level field. How far can he throw the same ball vertically upward? Assume that his muscles give the ball the same speed in each case.
Answer:
Using the range equation
R = vi2 sin (2θi)/g
the maximum range occurs when θi = 450 , and has a value
R = vi2 sin (900)/g = vi2/g
Given R, this yields
vi = √(gR)
If the boy uses the same speed to throw the ball vertically upward, then we use
vyf = vyi – gt
vyf = √(gR) – gt
At the maximum height, vyf = 0, giving
0 = √(gR) – gt
t = √(R/g)
and so the maximum height reached is
yf = yi + vyit + ½ ayt2
yf = 0 + t√(gR) + ½ (–g)t2
ymax = √(R/g) √(gR) + ½ (–g)(√(R/g))2
ymax = R/2
Problem#3
A quarterback throws a football straight toward a receiver with an initial speed of 20.0 m/s, at an angle of 30.0° above the horizontal. At that instant, the receiver is 20.0 m from the quarterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?
Answer:
The football travels a horizontal distance, we use
R = vi2 sin (2θi)/g
R = (20.0 m/s)2 sin (600)/(9.80 m/s2) = 35.3 m
Time of flight of ball is
t = (2vi sin θi)/g
t = 2(20.0 m/s) sin (300)/(9.80 m/s2) = 2.04 s
The receiver is ∆x away from where the ball lands and ∆x = 35.3 m – 20.0 m = 15.3 m. To cover this distance in 2.04 s, he travels with a velocity
∆x = vxt
vx = ∆x/t = 15.3/2.04 s = 7.50 m/s in the direction the ball was thrown
Problem#4
Your grandfather is copilot of a bomber, flying horizontally over level terrain, with a speed of 275 m/s relative to the ground, at an altitude of 3000 m. (a) The bombardier releases one bomb. How far will it travel horizontally between its release and its impact on the ground? Neglect the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before
he can call, “Bombs away!” Consequently, the pilot maintains the plane’s original course, altitude, and speed through a storm of flak. Where will the plane be when the bomb hits the ground? (c) The plane has a telescopic bomb sight set so that the bomb hits the target seen in the sight at the time of release. At what angle from the vertical was the bomb sight set?
Answer:
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| Fig.2 |
(a) Given: if we take the floor reference point then, yi = 3000 m, yf = 0, vxi = 275 m/s and vyi = 0 (because the plane initially moves horizontally)
Then, we use
∆x = vxit
then t = x/275
and yf = yi + vyit – ½ gt2, then
0 = 3000 + 0 – ½ (9.80 m/s2)t2
3000 = 4.9(x/275)2
x = 6800 m = 6.80 km
(b) The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be 3 000 m directly above the bomb when it hits the ground.
(c) When φ is measured from the vertical, tan φ = ∆x/∆y
therefore,
φ = tan-1 [∆x/∆y] = tan-1[6800/3000] = 66.20.
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