Falling Body: thrown upward w/ initial v=5 ft/sec from 40 ft
- Thread starter kcbrat
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D
Deleted member 4993
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Re: Falling Body
Hint:
First find out how long (time and distance) it will take for the upward speed to become zero.
That is maximum height.
Then it has to travel downward that much distance plus 40 feet.
Please share with us your work/thoughts - so that we know where to begin to help you.kcbrat said:
Hint:
First find out how long (time and distance) it will take for the upward speed to become zero.
That is maximum height.
Then it has to travel downward that much distance plus 40 feet.
D
Deleted member 4993
Guest
Re: Falling Body
Tell us interpretation of your answer - what did you find?
kcbrat said:
Tell us interpretation of your answer - what did you find?
Re: Falling Body
We can use the y component and solve for t.
Let the point from which the rock is thrown is the origin, (0,0).
Therefore, it hits the ground when y= -40
\(\displaystyle y=v_{0}t-\frac{1}{2}gt^{2}\)
\(\displaystyle -40=5t-16t^{2}\)
Solve for t.
We can use the y component and solve for t.
Let the point from which the rock is thrown is the origin, (0,0).
Therefore, it hits the ground when y= -40
\(\displaystyle y=v_{0}t-\frac{1}{2}gt^{2}\)
\(\displaystyle -40=5t-16t^{2}\)
Solve for t.
D
Deleted member 4993
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Re: Falling Body
Try to understand the formula you are using - you got 1.74 by setting your height to (-40).
If you have calculated correctly - then at 1.74 seconds the height is -40 - not maximum.
This is an equation of parabola. You know two points where it crosses the zero line. Where would be the vertex (max. height) of the parabola?
kcbrat said:ooh! 1.7451 is the time before it hits the ground.now, for the maximum height, do i just plug that in? or is there another formule?
Try to understand the formula you are using - you got 1.74 by setting your height to (-40).
If you have calculated correctly - then at 1.74 seconds the height is -40 - not maximum.
This is an equation of parabola. You know two points where it crosses the zero line. Where would be the vertex (max. height) of the parabola?
A body is thrown straight up with initial velocity 5 feet per second from a height of 40 feet. After how many seconds will it hit the ground? What will be its maximum height?
Time to maximum height above launch point derives from
Vf = Vo - gt where 0 = 5 - 32t making t(up) = .15625 sec.
The height reached derives from h = Vot - 16t^2 = 5(.15625) - 16(.15625)^2 making h = .39 feet.
In the same time period of .15625 sec., it will fall back to the launch height attaining a downward speed of 5 ft./sec.
The time to reach the ground from this point derives from 40 = 5t + 16t^2 from which t = 1.43 sec.
The total time from upward launch at 5 ft./sec. to impact is therefore 2(.15625) + 1.43 = 1.74 sec.
The maximum height reached from the ground is .39 + 40 = 40.39 ft.
Time to maximum height above launch point derives from
Vf = Vo - gt where 0 = 5 - 32t making t(up) = .15625 sec.
The height reached derives from h = Vot - 16t^2 = 5(.15625) - 16(.15625)^2 making h = .39 feet.
In the same time period of .15625 sec., it will fall back to the launch height attaining a downward speed of 5 ft./sec.
The time to reach the ground from this point derives from 40 = 5t + 16t^2 from which t = 1.43 sec.
The total time from upward launch at 5 ft./sec. to impact is therefore 2(.15625) + 1.43 = 1.74 sec.
The maximum height reached from the ground is .39 + 40 = 40.39 ft.