A stone is thrown vertically upwards with a velocity of 20 ms-'. Find the
maximum height reached by the stone and the total time of flight.
Answers
Given:-
- Initial velocity = u = 20 m/s
To find:-
- Maximum height reached by the stone
- Total time of flight
Answer:-
▪Maximum height reached by the stone:
Considering the motion from the instant the stone is projected upwards and the instant it reaches maximum height,
and using the formula,
- v² = u² + 2as
where,
- v is final velocity
- u is initial velocity
- a is acceleration
- s is Displacement
v will be 0 m/s, since the stone momentarily comes at rest at the highest point.
[Sign convention]: u, and s will be +ve as they are in upward direction, whereas a will be -ve (-9.8 m/s²) as it is downward direction.
Putting the values given,
0² = 20² + 2(-9.8)s
→ 0 = 400 - 19.6s
→ 400 = 19.6s
→ s = 20.41 m Ans
▪Time of flight:
We know that, time of flight = time of ascent + time of descent.
In this case, time of ascent = time of descent.
→ Time of flight = 2 * time of ascent
So, first we have to find time of ascent.
Considering the motion from the instant the stone is projected upwards and the instant it reaches maximum height,
and using the formula,
- v = u + at
where,
- v is final velocity
- u is initial velocity
- a is acceleration
- t is time of ascent
v will be 0 m/s, since the stone momentarily comes at rest at the highest point.
[Sign convention]: u will be +ve as it is in upward direction, and a will be -ve (-9.8 m/s²) as it is in downward direction.
Putting the values given,
0 = 20 + (-9.8)t
→ 9.8t = 20
→ t = 2.041 s
So, time of flight = 2 * time of ascent
→ time of flight = 2 * 2.041 s
→ time of flight = 4.082 s Ans