1. A train moving at a constant speed of 52.0 km/h moves east for 38.0 min, then

Question

1. A train moving at a constant speed of 52.0 km/h moves east for 38.0 min, then in a direction 45.0° east of due north for 21.0 min, and then west for 65.0 min. What is the average velocity of the train during this run? a) magnitude- km/h b) ° (counterclockwise from east) 2. A golf ball is struck at ground level. The speed of the golf ball as a function of time is shown in the figure below, where t = 0 at the instant the ball is struck. The graph is marked in increments of 0.25 s along the time axis, and vmin = 18.60 m/s and vmax = 22.27 m/s. (Values in figure do not necessarily match values in problem). a) How far does the golf ball travel horizontally before returning to ground level? b) What is the maximum height above the ground level attained by the ball? 3. A projectile's launch speed is 3.1 times its speed at maximum height. Find launch angle ?0. 4. In the figure, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.37 s. later, at distance d = 20.5 m from the building and at angle ? = 56° with the horizontal. a) Find h= m b) What is the velocity at which the ball is thrown? Magnitude= m/s Angle= ° relative to the horizontal c) Is that angle above or below the horizontal? Above/below

Explanation / Answer

1)

the displacement to the east = d1 = 52.0*(38.0/60) +52.0*(21.0/60)*sin(45.0) - 52.0*(65.0/60) = -10.53 km
the displacement to the north = d2 =52.0*(21.0/60)*cos(45.0) = 12.87 km
magnitude of total displacement d =?(d12 + d22) =16.63 km
total time t = (38.0 + 21.0 + 65.0)/60 hr = 2.07 hr
magnitude of the average velocity v = d/t= 16.63/2.07 = 8.034 km/h
direction ? =180o - tan-1(12.87/10.53) =129.29o

2)

a)

Vx = 18.60 m/s

Vy = sqrt[22.27^2 - 18.60^2]
Vy = 12.25 m/sec

d = Vy(t) + (1/2)(g)t^2
0 = 12.25t - 4.9t^2
4.9t^2 - 12.25t = 0
t(4.9t - 12.25) = 0
t = 0
4.9t - 12.25 = 0
t = 12.25/4.9
t = 2.5 sec

dx = 18.60 m/s*(2.5 sec) = 46.5 m
b)

2gdmax = Vf^2 - Vy^2
2(-9.8)(dmax) = - (12.25)^2
dmax = - (12.25)^2/2(-9.8)
dmax = 7.66 m

3)

Let

V = launch speed

V(cos ?) = horizontal component of the speed at its maximum height

? = launch angle

Since

V = 3.1V(cos ?)

then

cos ? = 1/3.1 = 0.3226

? = arc cos 0.3226

? = 71.18 degrees

4)

Vx = 20.5 / 1.37 = 14.96 m/s the constant horizontal speed
tan 56 = Vyf / Vx
Vyf = 14.96 * tan 56 = 10.09 m/s the final vertical speed
-Vyf = Vyi - g t taking upward direction as positive (assuming thrown upwards)
Vyi = 9.8 * 1.37 - 10.09 = 3.34 m/s (thrown upwards)
h = Vyi * t - 1/2 g t^2 = 3.34 * 1.37 - 1/2 * 9.8 * 1.37^2 = -4.62 m
Height = 4.6 m
Tan theta = 3.34 / 14.96 = .221 Theta = 12.48 deg

c)

above horizontal

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.