A stone is dropped from a balloon at an altitude of 300m. how long will the stone take to reach the ground if (a) the balloon is ascending with a velocity of 5m/s. (b) the balloon is descending with a velocity of 5m/s. (c) the balloon is stationary?
Answers
Explanation:
A stone is dropped from a balloon at an altitude of 300 m
a) how long the stone will reach the ground if a balloon is ascending with a velocity of 5 m
Initial velocity of stone upward u= 5m/s
By using equation of motion :
s=ut+\frac{1}{2}gt^2s=ut+
2
1
gt
2
Here displacement is in negative y direction and initially it was going upwards s=-300
g=-10m/s^2g=−10m/s
2
So,-300=5t+\frac{1}{2}(-10)t^2−300=5t+
2
1
(−10)t
2
\begin{gathered}-300=5t-5t^2\\-60=t-t^2\\t^2-t-60=0\\t=\frac{1}{2}-\frac{\sqrt{241}}{2},\frac{1}{2}+\frac{\sqrt{241}}{2}\\t=-7.262,8.26\end{gathered}
−300=5t−5t
2
−60=t−t
2
t
2
−t−60=0
t=
2
1
−
2
241
,
2
1
+
2
241
t=−7.262,8.26
Since time cannot be negative
Hence the stone will reach the ground in 8.26 seconds
b) how long the stone will reach the ground if a balloon is descending with a velocity of 5 m
u = 5 m/s
g = 10 m/s^2g=10m/s
2
s=300
Using equation of motion
s=ut+\frac{1}{2}gt^2s=ut+
2
1
gt
2
\begin{gathered}300=5t+\frac{1}{2}(10)t^2\\300=5t+5t^2\\60=t+t^2\\t^2+t-60=0\\t=-\frac{1}{2}-\frac{\sqrt{241}}{2},-\frac{1}{2}+\frac{\sqrt{241}}{2}\\t=−8.26,7.26\end{gathered}
300=5t+
2
1
(10)t
2
300=5t+5t
2
60=t+t
2
t
2
+t−60=0
t=−
2
1
−
2
241
,−
2
1
+
2
241
t=−8.26,7.26
Since time cannot be negative
Hence the stone will reach the ground in 7.26 seconds