A rocket is launched to travel vertically upwards with a constant velocity of 20m/s. after travelling for 35s the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. What is the total height achieved by it.? After what time of its launch does it come back to earth?
Answers
Hi, this is already answered by Santy2: https://brainly.in/question/5611117
S = ut
This is because it is moving at a constant velocity.
After 35 seconds the distance will be :
S = 20 × 35 = 700m
Now from 700 m above the ground now it travels at a free fall.
The initial velocity = 20 m/s
Final velocity = 0
Acceleration due to gravity = 10
Now :
V² = U² - 2gs
0 = 400 - 20S
20s = 400
S = 400/20
S = 20 m
The total height achieved = 700 + 20
= 720 m
Time it takes to drop back to the earth is :
S = ut + 0.5gt²
u = 0
720 = 0.5 × 10 × t²
720 = 5t²
t² = 720/5
t² = 144
t = √144
= 12 seconds
The total time it takes is :
12 + 35 = 47 seconds
S = ut
This is because it is moving at a constant velocity.
After 35 seconds the distance will be :
S = 20 × 35 = 700m
Now from 700 m above the ground now it travels at a free fall.
The initial velocity = 20 m/s
Final velocity = 0
Acceleration due to gravity = 10
Now :
V² = U² - 2gs
0 = 400 - 20S
20s = 400
S = 400/20
S = 20 m
The total height achieved = 700 + 20
= 720 m
The time it takes to drop back to the earth is :
S = ut + 0.5gt²
u = 0
720 = 0.5 × 10 × t²
720 = 5t²
t² = 720/5
t² = 144
t = √144
= 12 seconds
The total time it takes is :
12 + 35 = 47 seconds