2 D Motion Problems Solutions

Physics 110

Spring 2006

2-D Motion Problems: Projectile Motion – Their Solutions

1. A place-kicker must kick a football from a point 36 m (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.1m high. When kicked the all leaves the ground with a speed of 20 m/s at an angle of 530 to the horizontal.

1. Does the ball clear or fall short of the crossbar?
2. Does the ball approach the crossbar while still rising or while falling?

(a) From our equations of motion, the horizontal velocity is constant. This gives us the flight time for any horizontal distance starting with initial x velocity vicosθ. Thus the vertical height of gx 2 the trajectory is given as y = x tan θi – . With x = 36.0 m, vi = 20.0 m/s, and θ =

2v 2i cos2 θi
53.0°, we find y = (36.0 m)(tan 53.0°) –
(9.80 m/s2)(36.0 m)2

= 3.94 m. The ball clears the

(2)(20.0 m/s)2 cos2 53.0°

bar by (3.94 – 3.05) m =
0.889 m
.

(b) The time the ball takes to reach the maximum height is t
= vi sin qi

=

(20.0 m/s)(sin 53.0°)

1

g

x

= 1.63 s. The time to travel 36.0 m horizontally is t

=

, which gives t =

2

vix

9.80 m/s2

2

36.0 m
= 2.99 s.

(20.0 m/s)(cos 53.0°)

Since t2 > t 1 the ball clears the goal on its way down .

2. A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8m/s at an angle of 200 below the horizontal, where it strikes the ground 3 seconds later.

1. How far horizontally from the base of the building does the ball strike the ground?
2. At what height was the ball thrown?
3. How long does it take the ball to reach a point 10m below the level of launching?

(1) x = vxit = (8.00 cos 20.0°)(3.00) = 22.6 m

(b) Taking y positive downwards, y
= vyit +
1
gt2 = 8.00(cos 20.0°)3.00 +

2

1
(9.80)(3.00) 2 =

52.3 m

2

1
(c) 10.0 = 8.00 cos 20.0° t + 2 (9.80) t2, which gives a quadratic in t. The solutions to