a stone is thrown in a vertically upward direction with a velocity of 10m/s if the acceleration of the stone during its motion is 10m/s^2 in the downward direction what will be the height attained by the stone and how much time will it take to reach there?
Answers
Given :
▪ Initial velocity of ball = 10m/s
▪ Acc. due to gravity = 10m/s²
To Find :
➳ Max. height attained by ball.
➳ Time taken by ball to reach at maximum height.
SoluTion :
↗ Acc. due to gravity continuously acts in downward direction throughout the motion.
↗ For a body thrown vertically upward, g is taken negative.
↗ Final velocity of ball at max. height = 0
✴ Max. height attained by ball :
⇒ v² - u² = 2(-g)H
⇒ 0² - 10² = 2(-10)H
⇒ 0 - 100 = -20H
⇒ H = -100/-20
⇒ H = 5m
✴ Time taken by ball to reach at there :
➨ v = u - gt
➨ 0 = 10 - 10t
➨ t = 10/10
➨ t = 1s
Given :-
- Initial Velocity (u) = 10m/s
- Acceleration due to gravity (g) = 10m/s² .
To Find :-
- What will be the height attained by the stone ?
- And how much time will it take to reach there ?
Solution :-
Condition (1)
Maximum height attained by ball.
Applying third eqn. of motion, we get
=> v² = u² + 2(-g)s
=> 0² = 10² + 2( - 10) × s
=> 0 - 100 = -20 × s
=> -100 = -20 × s
=> s = -100/-20
.°. H = 5m ( °.° S = H )
Condition (2)
The time will it take to reach there .
Applying first eqn. of motion , we get
=> v = u - gt
=> 0 = 10 - 10 × t
=> -10 = -10 × t
=> t = 10/10
=> t = 1sec