 # Solutions to Practice Questions from Module 5:

```Solutions to Practice Questions from Module 5:
1. A golf ball is hit with a velocity of 24.5 m/s at 35.0 degrees above the horizontal. Find
a) the range of the ball, and
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 24.5 m/s @ 35.0°
y=0m
x=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v x = vi cos q = (24.5m / s)(cos 35) = 20.0692m / s
viy = vi sin q = (24.5m / s)(sin35) = 14.0526m / s
†
Need to determine the time it took to land:
y = viy t + 21 at 2
0m = (14.0526m / s)t + 21 (-9.81m / s2 )t 2
0m = (14.0526 / s - 4.905m / s2 t)t
t = 0s
0m = 14.0526 / s - 4.905m / s2 t
14.0526m / s
t=
4.905m / s2
t = 2.8649s
x = vxt
†
x = (20.0692m / s)(2.8649s)
x = 59.4963m
The range of the ball is 59.5 meters.
† maximum height of the ball.
b) the
Half the flight time will be 1.4324 seconds.
y = viy t + 21 at 2
y = (14.0526m / s)(1.4324s) + 21 (-9.81m / s2 )(1.4324)2
y = 20.1289m -10.0639m
y = 10.0650m
The ball will reach a maximum height of 10.1 meters.
†
2. A player kicks a football from ground level at 27.0 m/s at an angle of 30.0 degrees above
the horizontal. Find
a) its "hang time" (time that the ball is in the air),
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 27.0 m/s @ 30.0°
y=0m
t=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v x = vi cos q = (27m / s)(cos 30) = 23.3827m / s
viy = vi sin q = (27m / s)(sin 30) = 13.5m / s
y = viy t + 21 at 2
†
0m = (13.5m / s)t + 21 (-9.81m / s2 )t 2
0m = (13.5 / s - 4.905m / s2 t)t
t = 0s
0m = 13.5 / s - 4.905m / s2 t
13.5m / s
t=
4.905m / s2
t = 2.7523s
The ball was in the air for 2.75 seconds.
† distance the ball travels before it hits the ground, and
b) the
x = vxt
x = (23.3727m / s)(2.7523s)
x = 64.3287m
The ball traveled 64.3 meters
c) its†maximum height.
Half the flight time will be 1.3762 seconds.
y = viy t + 21 at 2
y = (13.5m / s)(1.3762s) + 21 (-9.81m / s2 )(1.3762)2
y = 18.5787m - 9.2697m
y = 9.3090m
The ball will reach a maximum height of 9.31 meters.
† retrieving her ball from the roof, a girl throws the ball off of the roof with a speed of
3. After
5.6 m/s at an angle of 35 degrees above the horizontal. The ball is thrown from a height of
11 m above the ground. Find
a) the time the ball is in the air,
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 5.6 m/s @ 35°
y = -11 m
x=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v x = vi cos q = (5.6m / s)(cos 35) = 4.5873m / s
viy = vi sin q = (5.6m / s)(sin 35) = 3.2120m / s
y = viy t + 21 at 2
†
-11m = (3.2120m / s)t + 21 (-9.81m / s2 )t 2
(4.905m / s2 )t 2 - (3.2120m / s)t -11m = 0
†
t=
=
-b ± b2 - 4ac
2a
-(-3.2120) ± (-3.2120)2 - 4(4.905)(-11)
2(4.905)
3.2120 ± 10.3169 + 215.82
9.81
3.2120 ± 226.1369 3.2120 ± 15.0378
=
=
9.81
9.81
3.2120 + 15.0378 3.2120 -15.0378
=
or
9.81
9.81
18.2498 -11.8258
=
or
9.81
9.81
= 1.8603s or -1.2055s
=
†
The ball will land after 1.9 seconds.
b) the range of the ball,
Solution:
x = vxt
x = (4.5873m / s)(1.8603s)
x = 8.5338m
The ball will travel 8.5 meters away from the base of the building.
† maximum height of the ball, and
c) the
Solution:
In this case (because the take-off and landing heights are not the same) the
maximum height will not occur at the half-way time. We do however, know that
the maximum height occurs when vy = 0 m/s.
Find the distance when vy = 0 m/s
2
2
v fy = viy + 2ay
(0m / s)2 = (3.2120m / s)2 + 2(-9.81m / s2 )y
0 = 10.3169m2 / s2 - (19.62m / s2 )y
10.3169m2 / s2
19.62m / s2
y = 0.5258m
y=
†
This gives us the distance above the starting point. But since the starting point
is 11 meters off the ground we also need to take that into account.
The ball reaches a maximum height of 11.5 meters above the ground.
d) the speed of the ball as it strikes the ground.
Because vx doesn’t change we know the horizontally velocity of the ball remains
constant at 4.5873 m/s.
Need to determine the final velocity in the vertical direction.
v fy = viy + at
v fy = 3.2120m / s + (-9.81m / s2 )(1.8603s)
v fy = 3.2120m / s -18.2495
v fy = -15.0375
†
To determine the final speed we need to resolve the perpendicular vectors.
2
2
v f = v fx + v fy
2
2
v f = (4.5873m / s)2 + (-15.0375m / s)2
2
v f = 21.0433m2 / s2 + 226.1264m2 / s2
2
v f = 247.1697m2 / s2
v f = 247.1697m2 / s2
v f = 15.7216m / s
The ball hits the ground with a speed of 15.7 m/s.
†
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