Question

If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?

Answer 1

Answer:

$heigth=2.86m$

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

$V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s$

Also we know that average velocity

$V_{avg}=(V_{i}+V_{f})/2$

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

$V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862$

Substitute value of Vf in average velocity

So

$V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s$

Vi is speed of balloon at top of the window

Now we need to find time

So

$V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s$

So the distance can be found as

$distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m$