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Two-Dimensional Motion

  • Thread starter frozen7
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  • #1
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A stone is thrown upward from the top of a building at an angle of 30 degree to the horizontal and with an initial speed of 20m/s. The height of the building is 45m.
How long is the stone "in flight"?

Solution:

Vx0 = 20cos30 = 17.3
Vy0 = 20sin30 = 10

-45 = 10t - 1/2 g t^2
t = 4.22s

This is an example from my physics textbook. I wonder why the solution is in this way. Why the time used to reach the highest level and the time from the highest level to the initial position are not considered as time "in flight"?

Thanks.
 

Answers and Replies

  • #2
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frozen7 said:
A stone is thrown upward from the top of a building at an angle of 30 degree to the horizontal and with an initial speed of 20m/s. The height of the building is 45m.
How long is the stone "in flight"?

Solution:

Vx0 = 20cos30 = 17.3
Vy0 = 20sin30 = 10

-45 = 10t - 1/2 g t^2
t = 4.22s

This is an example from my physics textbook. I wonder why the solution is in this way. Why the time used to reach the highest level and the time from the highest level to the initial position are not considered as time "in flight"?

Thanks.
The solution is correct.
I do not get your question though. The time to reach maximal height is also considered to be time in flight. Why do you get this idea. Besides, the time to get to max height is not calculated here, you are just calculating the time it takes for the vertical y-coordinate to go from 45 to 0. That is all, and that is ofcourse the complete motion.
The general formula in this case is [tex]y=y_{initial} + v_{initial}t -gt^2/2[/tex] and y=0 here

marlon
 
Last edited:
  • #3
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Then what does "time in flight" actually mean?
Not from the time it is thrown until it reaches the ground?
 
  • #4
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frozen7 said:
Then what does "time in flight" actually mean?
Not from the time it is thrown until it reaches the ground?
yes the time it is thrown until it reaches the ground's surface

marlon
 
  • #5
HallsofIvy
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If you were to graph y= 10t - 1/2 g t2 the graph would be, of course, a parabola. (Since the x motion is linear, x= 17.3t, the actual path is a parabola.)

It starts at (0, 0) rises to some maximum, then falls to (4.22, -35). You may be thinking that you calculate the time going up, then the time going down and add but that is not necessary: solving y= -35= 10t- (1/2)gt2 gives the total time immediately. Notice that a quadratic equation has two solutions (the other solution with y= -35 is negative and so not a "time in flight"). If you took y positive but less than the maximum height, your quadratic equation would have two positive solutions: the smaller the time at which the stone passes that height going up, the larger, the time the stone passes that height on its way back down. What do you think would happen if you set y equal to a number greater than the maximum height and solved for t?
 
  • #6
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HallsofIvy said:
If you were to graph y= 10t - 1/2 g t2 the graph would be, of course, a parabola. (Since the x motion is linear, x= 17.3t, the actual path is a parabola.)
This is not the correct formula.
It is : [tex]y=45 +10t -gt^2/2[/tex]
It starts at (0, 0) rises to some maximum,
If you chose the origin just at the bottom of the building, this get easier in my opinion. The above formula is based upon this assumption

marlon
 
  • #7
mukundpa
Homework Helper
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The time of flight is simply the time for which the object remains in air and moving under gravity. Here it is from the top of the building to the ground and the time to reach the maximum height is not equal to the time from max height to the ground. For ground to ground both distances are equal. I think Frozen 7 required that much only. Is it?
 

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