As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packa

Question

As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure 2-31 gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it rise? (b) How high is the break-free point above the ground?

I have tried this problem several times and cannot get it...if someone could please help me out it would be greatly appreciated!!!

Explanation / Answer

The package breaks free at 2s because this is when it starts slowing down and stops ascending at 4s since that is when its velocity reaches 0m/s so it rises for 2s.

To find the distance it travels
y =voyt + 1/2 gt2

y = 19.6m/s * 2s + 1/2 (-9.8)22 19.6m above the break away point.

Then at 4s, the package is at its highest point and then falls for 4s until it reaches the ground. You can use the same equation to find the distance that it falls but with no initial velocity.

y = 1/2 gt2 = 1/2 *9.8 *42 = 78.4m - this is total distance traveled. You need to subtract the distance after the breakaway point to see how high the break away point is.

78.4m - 19.6m = 58.8m

Hope that helps

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