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Projectile Motion - Horizontal launches

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Projectile Motion - Horizontal launches The scale on the left is 1cm = 5m Draw the positions of the dropped ball every second that it is in the air. – PowerPoint PPT presentation

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Title: Projectile Motion - Horizontal launches


1
Projectile Motion - Horizontal launches
The scale on the left is 1cm 5m Draw the
positions of the dropped ball every second that
it is in the air. Neglect air resistance and
assume g 10 m/s2. Estimate the number of
seconds that the ball is in the air
Four positions of a ball thrown horizontally
without gravity are shown. Draw the positions of
the ball with gravity. Describe the path taken by
the ball.
It is parabolic
How is motion in the vertical direction affected
by motion in the horizontal direction?
They are independent of each other
2
Projectile Motion
Rules
1. Vertical Motion is effected by gravity
2. Horizontal motion is constant
  1. Motion in the horizontal direction is independent
    of the motion in the vertical direction.

4. Assume no air resistance
3
Physical Expressions
Horizontal
Vertical
vHi vHf vH
vvf vVi g ?t
?dH vH ?t
?dv vvi ?t 1/2 g ?t2
a 0 m/s2
a g -9.8 m/s2
4
Example A Falling Care Package
The airplane is moving horizontally with a
constant velocity of 115 m/s at an altitude of
1050m. Determine the time required for the care
package to hit the ground.
5
Example A Falling Care Package
?dv av vvf vvi ?tair

-1050 m
- 9.8 m/s2
0 m/s
?
6
Example A Falling Care Package
?dv g vvf vvi ?tair
-1050 m -9.80 m/s2 0 m/s ?
7
Example A Falling Care Package
What is the horizontal distance traveled by the
care package?
?dH
8
Example A Falling Care Package
?dH
?dH aH vH ?tair

?
0 m/s2
115 m/s
14.6 s
9
Example A Falling Care Package
?dH aH vH ?tair
? 0 m/s2 115 m/s 14.6 s
10
Example Problem
Supposed that you are an accident investigator
and are asked to figure out whether the car was
speeding before it crashed through the rail of
the bridge into the mudbank as shown below. The
speed limit on the bridge is 55 mph (24m/s).
Whats your conclusion?
?
vvi 0 m/s ?dH 24m ?dv - 4.9 m g -
9.8 m/s2 vH ?
?dv vvi ?t 1/2 g ?t2
- 4.9 m
- 4.9 m (0 m/s) ?t 1/2 (-9.8 m/s2) ?t2
?t ? (2 (-4.9 m) / -9.8 m/s2) 1 s
?dH vH ?t
24 m
24m vH (1s)
So vH 24m / 1s 24 m/s
The car loses speed when it went through the
railing so it must have been speeding
11
Example Continued
vH
?
vvf
What speed did the car hit the mudbank?
vBottom
Easiest to solve using conservation of energy
vTop 24 m/s hTop 4.9 m hBottom 0m g
-9.8 m/s2 vBottom ?
KETop GPETop KEBottom GPEBottom
1/2 m vTop2 (-mghTop) 1/2 m vBottom2
(-mghBottom)
1/2 vTop2 (-ghTop) 1/2 vBottom2
(-ghBottom)
1/2 (24 m/s)2 (-(-9.8 m/s2) (4.9m) 1/2
vBottom2 (-(-9.8m/s2)(0m)
336 m 1/2 vBottom2
vBottom ? 2(336 m) 26 m/s
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