Practice 3D: | 1 | 2 |
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Projectile Motion - by Kevin Pruyn,
2003
1. An autographed baseball rolls off of a 0.70m high desk and strikes the floor 0.25m away from the base of the desk. How fast was it rolling?
For starters, you need to set up the handy-dandy "SUVAT" table. We know that the vertical displacement is 0.70m high, and the horizontal displacement is 0.25m. What they don't tell us but what we must assume is that the initial vertical velocity is 0, because the ball is being rolled across a table, and the acceleration is -9.8m/s/s for gravity. We also know that since there is know horizontal acceleration, it has to be 0. So now our table looks like this...
H V Dx 0.25m 0.70m vi ? 0 m/s vf a 0m/s/s -9.8m/s/s t So now we can use the equation vf2 = vi2 + 2a(Dx) to solve for vf2
vf2 = 02 + 2(-9.8)(0.70)
vf2=√13.72
vf=3.704051835m/s
(note disregard the negative sign under the square root. We decided to make it negative so we can magically turn it back to positive).
Now with just time left on the right side of our table we can solve for time using
Dx = 1/2(vi + vf)Dt
0.7=1/2(3.704+0)Dt
Dt=.37796 seconds
Now that we found time and we know that "time is on both sides" our little table looks like this...
H V Dx 0.25m 0.70m vi ? 0 m/s vf 3.704m/s a 0m/s/s -9.8m/s/s t 0.37796 sec 0.37796 sec Now with only the initial horizontal velocity left to find we can go ahead and find the answer using...
Dx = vi(Dt) + 1/2a(Dt)2
0.25=vi(.37796) + 1/2(0)(9.37796)2
0.25=vi(.37796)
vi=.661445 m/s
2. A cat chases a mouse across a 1.0m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2m from the edge of the table. What was the the cat's speed when it slid off the table?
This is the exact same problem as number one, so I don't think I have to explain myself. First you find the vertical final velocity using vf2 = vi2 + 2a(Dx). You should get 4.427m/s. Then you solve for time by using Dx = 1/2(vi + vf)Dt. Your time should be about .451773 sec. Then lastly use Dx = vi(Dt) + 1/2a(Dt)2 to find our final horizontal velocity which should be
4.87 m/s.
3. A pelican flying along a horizontal path drops a fish from a height of 5.4m. The fish travels 8.0 m horizontally before it hits the water below. What is the pelican's initial speed?
This problem is very similar to number 1 and 2. Again we will set up our table to look something like this.( Remember that horizontally, the acceleration is zero and vertically in -9.8. Also the vertical initial velocity is zero).
H V Dx 8.0m 5.4m vi ? 0 m/s vf a 0m/s/s -9.8m/s/s t Use vf2 = vi2 + 2a(Dx) to find the final vertical velocity. You should get 10.28785692m/s. Next find time using Dx = 1/2(vi + vf)Dt. The answer should come out to 1.0497086242sec. All that is left is to use Dx = vi(Dt) + 1/2a(Dt)2 to find the initial speed of the fish. The answer comes out to be
7.62m/s
4. If the pelican in item 3 was traveling at the same spee but was only 2.7m above the water, how far would the fish travel horizontally before hitting the water below.?
For this problem we are going to take the same approach but in the end we will be solving for different variables. So once we set up our table again it should look like this...
H V Dx ? 5.4m vi 7.62m/s 0 m/s vf a 0m/s/s -9.8m/s/s t Notice, however, that the question mark is in the horizontal distance this time. This is because this is the answer we are trying to find if the pelican's intial speed is 7.62m/s (the answer we found to the last problem). So using the same equations
vf2 = vi2 + 2a(Dx)
Dx = 1/2(vi + vf)Dt
Dx = vi(Dt) + 1/2a(Dt)2
we can then find the horizontal displacement. Using the first equation you should a vertical final velocity of 7.2746m/s. Then use he second equation to find time. You should acquire an answer of .742304 sec. And lastly use the last equation to find the displacement. Since we are solving for Dx it should be a little easier to solve, just solve it out left to right. The answer is
5.66m