The object dropped from a height (h) with initial velocity zero strikes the ground with a velocity of 30m's.
How long does it take to reach the ground. Also find h lg = 10m/s)
Answers
Answered by
56
Given -
- Initial velocity (u) = 0 m/s
- Final velocity (v) = 30 m/s
- Acceleration due to gravity (g) =
To find -
- Time the object will take to reach the ground.
- Height.
Solution -
- v = u + at
Substitute the given values -
↬ 30 = 0 + 10 × t
↬ 30 = 10 × t
↬ = t
↬ t = 3s
Substitute the given values.
↬
↬
↬
↬ s = 45 m
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Know more -
Velocity -
- It is distance travelled by a body in one second in a particular direction.
- It is a vector quantity.
- Its S.I. unit is m/s.
Acceleration -
- It is the rate of change of velocity of a body.
- It is vector quantity.
- Its S.I. unit is
Equations of motion -
First equation of motion -
- v = u + at
Second equation of motion -
- s = ut +
Third equation of motion -
where,
v = Finial velocity
u = Initial velocity
a = acceleration
s = distance/displacement
t = time
Answered by
38
Given :
- Initial velocity, u = 0 m/s
- Final velocity, v = 30 m/s
- Acceleration due to gravity, g = 10 m/s
To find :
- Time taken to reach the ground (t)
- Height (h)
According to the question,
Note : Here,
- Acceleration (a) = Acceleration due to gravity (g)
→ v = u + at
Where,
- v = Final velocity
- u = Initial velocity
- a = Acceleration
- t = Time
→ Substituting the values,
→ 30 = 0 + 10 × t
→ 30 - 0 = 10t
→ 30 = 10t
→ 30 ÷ 10 = t
→ 3 = t
So,the time is 3 seconds.
Now,
→ v² = u² + 2as
Where,
- v = Final velocity
- u = Initial velocity
- a = Acceleration
- s = Distance
→ Substituting the values,
→ (30)² = (0)² + 2 × 10 × s
→ 900 = 0 + 20s
→ 900 - 0 = 20s
→ 900 = 20s
→ 900 ÷ 20 = s
→ 45 = s
So,the height is 45 metres.
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