Giambattista College Physics (3rd ED). Chapter 4, Problem 44. A suspension bridg
ID: 1963048 • Letter: G
Question
Giambattista College Physics (3rd ED). Chapter 4, Problem 44.A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistance. At the location of the bridge g has been measured to be 9.83 m/s^2
(a) If you drop the stone, how long does it take to fall to the base of the
gorge?
(b) If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground?
(c) If you throw the stone with a velocity of 20.0 m/s at an angle of 30.0° above the
horizontal, how far from the point directly below the bridge will it hit the level ground?
Explanation / Answer
A) s = ut + 1/2 at2
60 = 0 + 0.5 x 9.83 x t2
60 = 4.915 t2
t2 = 12.2075
t = 3.494 s
B) You can use the same equation as above but, unless you have a graphic calculator you would have to use the quadratic rule. So I'll show a method which will give you the answer with minimum fuss and chance of making a mistake
v2 = u2 + 2as
v2 = 202 + 2 x 9.83 x 60
v2 = 1579.6
v = 39.744 m/s/s
a = (v-u)/t
9.83 = (39.744 - 20)/t
t = 19.744/9.83
t = 2.009 s
C) First, find the components of the velocity
cos() = vx/v
cos 30 = vx / 20
vx = 20 x cos 30
= 17.32 m/s
sin() = vy/v
sin 30 = vy / 20
vy = 20 sin 30
= 10 m/s
Then calculate the time taken for it to hit the ground
a = (v-u)y/t1
- 9.83 = - 10 /t
t1 = 1.0173 s
s = ut + 1/2 at2
= 10 x 1.0173 + 1/2 x - 9.83 x 1.0173
= 10.173 - 5
= 5.173 m (+ 60)
s = ut + 1/2 at2
65.173 = 0 + 1/2 x 9.83 x t2
65.173 = 4.915t2
t2 = 13.26
t2 = 3.641 s
ttot = 3.641 + 1.0173
= 4.65873 s
sx = vxt
= 17.32 x 4.659
s = 80.689 m
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