A hot-air balloon has just lifted off and is rising at the constant rate of 1.8m/s . Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward.
If the passenger is 3.1m above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger? (Hint: When the camera is thrown with its minimum speed, its speed on reaching the passenger is the same as the speed of the passenger.)
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Let initial speed of the camera be u m/sec.Let further it reaches to passeneger t seconds after through .
Balloon will further rises (1.8 x t) meter.
Camera has to reach ballon by covering height of (3.1+1.8t) meter in t sec. .
Thus equation of linear motion under influence of gravity
`ut=h+(1/2) g t^2`
`` For minimum u ,
Thus t=.8 sec will give the minimum velocity of ballon.
Thus the minimum initial speed of the camera should be 9.6 m/s.
I included a quick sketch that helped me visualize the problem. The red line would be the balloon. The blue line would be the camera. The reason it is moved to the right is because of the timeframe. The graph plots the time (x) vs. the distance (y). And, the camera wasn't thrown up until after a certain amount of time. The black dot is where the camera would meet up with the balloon. So, we are looking for when those y's would be equal.
For the balloon, the equation for the height would be:
y = 1.8x
y = height of balloon
x = time balloon flies
The camera was thrown when the balloon was 3.1 meters high already. So, to find the time it was thrown, plug in 3.1 meters into this equation:
3.1 = 1.8x
Solving for x:
x = approx. 1.72 seconds after the balloon took off was the camera thrown.
So, now, the formula for the height of the camera would be:
y = -9.8x^2 + iv*x + id
y = height of camera
iv = initial velocity
id = initial distance
x = time camera flies
But, here, we need to include the 1.72 seconds the balloon has been in the air. To do that, we would subtract 1.72 from each x (time) in the equation. So:
y = -9.8(x-1.72)^2 + iv*(x-1.72) + id
This would also be how you would show a horizontal shift to the right, which would agree with our sketch earlier.
Also, we can assume id = 0, since the friend was on the ground at the time. So, the equation would be:
y = -9.8(x-1.72)^2 + iv*(x-1.72)
So, we have:
y = 1.8x for the balloon, and
y = -9.8(x-1.72)^2 + iv*(x-1.72) for the camera
But, we have 2 equations and 3 unknowns. So, we need one more equation.
The problem specifies that the velocity of the camera would be the same as the velocity of the balloon, which would match up with our sketch. So, the velocity of the camera is 1.8 m/s when it reaches the balloon. Also
v = iv - 9.8x
Where v = final velocity
iv = initial velocity
x = time
We know the final velocity has to be 1.8. So, we have:
1.8 = iv - 9.8x, no need to include the 1.72 sec here, since we are only interested with the time it takes the camera to achieve a velocity of 1.8 m/s, the flight of the camera.
So, now, we have our 3 equations and 3 unknowns. Taking this last one and solving for iv, we get iv = 9.8x + 1.8. Then, we could set the first two equations equal to each other:
1.8x = -9.8(x-1.72)^2 + iv*(x-1.72)
And, plug in the last equation for iv:
1.8x = -9.8(x-1.72)^2 + (9.8x + 1.8)*(x-1.72)
There are a variety of ways to solve this. I did the graphing method and found the intersection point on my graphing calculator. There, x = 1.90 second, the amount of time the camera would be in the air. So, with the last of our 3 equations:
1.8 = iv - 9.8x
And, we now know that x = 1.90, then, solving for iv:
iv = 9.8*1.90 + 1.8 = 20.42
So, the camera would have to be thrown with an initial velocity of at least 20.42 m/s for the balloon person to get the camera.
Good luck, nikki. I hope this helped.
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