Physics 106
Group Problems
Projectile Motion
Name:
Summer 2015
O
NS
TA:
1. (10 points) A mountain climber jumps a crevasse by leaping horizontally with speed
v0 = 5m/s. If the climber’s direction of motion on landing is θ = 40◦ below the horizontal:
(a) What is the height difference h between the two sides
of the crevasse?
Solution:
For the landing angle we have the relation:
vy
tan θ =
vx
Where vy and vx are the components of the velocity at the
moment of landing. Since there is no vertical component
of the initial velocity we can write:
TI
vy2 = −2g∆y = −2g(0 − h) = 2gh
p
vy = 2gh
vx = v0
Putting in the expressions for vx , vy , we get:
√
2gh
vo
2gh
tan2 θ = 2
vo
2
2
tan θv0
=h
2g
LU
tan θ =
h=
v02 tan2 θ
= 0.9m
2g
SO
(b) How far away does the climber move from his jump position?
Solution:
Time of the flight can be determined from the y component of the equation of motion
1
0 = h − gt2
2
s
t=
2h
= 0.43s
g
The displacement in horizontal direction is simply
R = v0 t
R = 2.14m
1
2.
O
NS
(10 points) A daredevil tries to jump a canyon of width R = 10m. To do so, he drives his
motorcycle up an incline sloped at an angle of θ0 = 15◦ as shown in the figure.
(a) What minimum launch speed is necessary to clear the canyon?
Solution:
Directly applying the equation for the range of a projectile
R=
r
v0 =
v02
sin 2θ0
g
g∆x
= 14m/s
sin 2θ0
TI
(b) How long it would take him to fly to the other side of the canyon?
Solution:
SO
LU
R = v0,x t = v0 cos(θ0 )t
R
t=
v0 cos θ0
10
= 0.74s
t=
14 cos 15◦
2
3.
O
NS
(10 points) A golfer hits the ball with a club from an elevated tee that lands on the green
close to the cup a distance of R = 165m from where he hit it as shown in the figure.
The spot where it lands is h = 5 m below
the spot where the ball is hit. The ball leaves
the club with a horizontal component of velocity
equal to v0,x = 50.0m/s. (Ignore air resistance)
s
(a) How long does the ball take to get to the
green?
Solution:
∆x = v0,x t
∆x
R
t=
=
v0c
v0,x
165
t=
= 3.3s
50
TI
(b) What is the vertical component of the ball’s initial velocity?
Solution:
The earth acceleration acts in the negative y direction and has magnitude of g = 9.8m/s2 . Also
taking into account that the difference y − y0 = −h the equation of motion in the y direction
becomes:
LU
1
y = y0 + v0,y t + at2
2
1
y − y0 = v0,y t − gt2
2
1
−h = v0,y t − gt2
2
−h 1
=
+ gt = 14.7m/s
t
2
SO
v0,y
3
4.
(10 points) A kitten jumps off a table and lands on the floor as shown in the figure. The
height of the table is h = 85cm. The kitten begins his jump with an initial speed of v0 = 5.0m/s‘
at an angle of θ0 = 53◦ above the horizontal.
O
NS
(a) Find the maximum height,hmax , that
the kitten attains.
Solution:
For this problem we will use the equation relating the initial, final velocities, the distance
traveled and acceleration.
2
2
vmax,y
− v0,y
= 2a∆y
At the peak of the projectile we have
• the y component of the velocity is zero vmax,y = 0
TI
• distance traveled in the y direction is ∆y = hmax − h
0 − (v0 sin θ0 )2 = −2g(hmax − h)
(v0 sin θ0 )2
hmax − h =
= 0.81m
2g
hmax = 1.7m
LU
(b) How far away does the kitten land from his starting point?
Solution:
R = v0,x t = vo t cos θ0
SO
Where t in the equation above is the flight time of the kitten that we need to find. The flight time
of the kitten can be derived from the y component equation of motion for the landing point y = 0
1
y = yo + v0,y t − gt2
2
1 2
0 = h + v0,y t − gt
2
The two solutions of the quadratic equation with respect to t are t1 = −0.18s and t2 = 0.99s.
q
1
2 + 2gh
t1,2 =
v0,y ± v0,y
g
The flight time of the kitten is 1 second. Using that for the range of the projectile we find
R = v0 cos θ0 × 1 = 3m
4
5.
(10 points) Prove that the landing speed of a projectile is independent of launch angle for
a given height of launch.
vy2
=
O
NS
Solution:
The components of the velocity at the moment of landing vy and vx are:
vx = v0 cos θ
2
v0 sin2 θ ± 2gh
Where h is the height from which the projectile was launched with respect to the landing point.
So the landing speed square is:
v 2 = vx2 + vy2 = v02 cos2 θ + v02 sin2 θ ± 2gh
v 2 = v02 (cos2 θ + sin2 θ) ± 2gh
v 2 = v02 ± 2gh
SO
LU
TI
The expression for the landing speed has no θ which means that it is independent of the launch
angle.
5
6.
(10 points) A ball is thrown at an angle θ0 above the horizontal. In the presence of the
acceleration due to gravity, how far below the straight-line path following its original direction of
motion will the ball be at the end of 1.0 second?
O
NS
Solution:
SO
LU
TI
1
yparab = y0 + v0,y t − gt2
2
yline = y0 + v0,y t
1
∆y = yline − yparab = gt2
2
∆y = 4.9m
6