A bottle dropped from a balloon reaches the ground in 20 s. Determine the height of the balloon if (a) it was at rest in the air and (b) it was ascending with a speed of 50 m/s when the ball was dropped. Ans. (a) 2.0 km; (b) 0.96 km We are
Answers
Step-by-step explanation:
given that , A bottle is dropped from a balloon which reached the ground in 20 seconds. We have to find the height at which the balloon was present if It was at rest in the air It was ascending with of a speed of 50m/s [1] Here, In this case, we have Time taken to reach ground = 20 s Initial velocity = 0 m/s Acceleration due to gravity, g = 10 m/s² Using the following formula, we get ⇒ T = v / g ⇒ 20 = v/10 ⇒ v = 200 m/s So, The final velocity of the bottle is 200 m/s and the initial velocity is 0. Using the third equation of motion,
we have ⇒ v² - u² = 2gs
⇒ 200² - 0 = 2×10×s
⇒ 40000 / 20 = s
⇒ s = 2000 m or 2 Km
[2] Here, It is given that the balloon is ascending at 50 m/s hence the initial velocity of the bottle is -50 m/s Total Time taken by the bottle to reach the ground = 20 seconds. Using the second equation of motion
, ⇒ s = ut + 1/2 gt²
⇒ s = ut + 1/2 gt²
⇒ s = -50 × 20 + 1/2 × 10 × 20²
⇒ s = - 1000 + 5 × 400
⇒ s = 2000 - 1000
⇒ s = 1000 m or 1 km.
Answer:
We are given that ,
A bottle is dropped from a balloon which reached the ground in 20 seconds.
We have to find the height at which the balloon was present if
It was at rest in the air
It was ascending with of a speed of 50m/s
[1]
Here, In this case, we have
Time taken to reach ground = 20 s
Initial velocity = 0 m/s
Acceleration due to gravity, g = 10 m/s²
Using the following formula, we get
⇒ T = v / g
⇒ 20 = v/10
⇒ v = 200 m/s
So, The final velocity of the bottle is 200 m/s and the initial velocity is 0.
Using the third equation of motion, we have
⇒ v² - u² = 2gs
⇒ 200² - 0 = 2×10×s
⇒ 40000 / 20 = s
⇒ s = 2000 m or 2 Km
[2]
Here, It is given that the balloon is ascending at 50 m/s hence the initial velocity of the bottle is -50 m/s
Total Time taken by the bottle to reach the ground = 20 seconds.
Using the second equation of motion,
⇒ s = ut + 1/2 gt²
⇒ s = ut + 1/2 gt²
⇒ s = -50 × 20 + 1/2 × 10 × 20²
⇒ s = - 1000 + 5 × 400
⇒ s = 2000 - 1000
⇒ s = 1000 m or 1 km.