Question

a stone is thrown in a vertically upward direction with a vertically upward direction with a velocity of 5ms. if the acceleration of the stone during its motion is 10ms. in the downward direction what will be the height attained by the stone and how much time will it take to reach there​

Answer 1

$\huge\underline{\overline{\mid{\bold{\orange{Given-}}\mid}}}$

• Initial velocity, u = 5 m/s
• Acceleration, a = - 10 m/s² (stone is coming downward)
• Final velocity, v = 0

$\huge\underline{\overline{\mid{\bold{\red{Formulas-}}\mid}}}$

By using equations of motion

• $\boxed{ v = u + at}$---(1)
• $\boxed{v^2 - u^2 = 2as}$----(2)

$\huge\underline{\overline{\mid{\bold{\blue{Solution-}}\mid}}}$

putting values in (1),we get

$0 = - 5 + (- 10) \times t$

$t= \dfrac{5}{10}= \dfrac{1}{2} = 0.5 seconds$

time= 0.5 sec

now putting values in (2), we get

$0^2- 5^2 = 2\times -10\times s$

$s=\dfrac{25}{20}= 1.25sec$

the height attained by stone is 1.25 seconds.

Answer 2

EXPLANATION.

• GIVEN

a stone is thrown in a vertically upwards

direction with a velocity = 5 m/s.

the acceleration of the stone during it's

motion = 10 m/s in downward direction.

1) =To find what will be the height attend

by the stone.

2) = How much time will it take to reach.

According to the question,

initial velocity = u = 5 m/s

acceleration = a = -g = 10 m/s².

From the Newton third equation of

kinematics.

=> v² = u² + 2as

=> (0)² = (5)² + 2(-10)s

=> 25 = 20s

=> s = 25/20

=> s = 1.25 m

From Newton first equation of

kinematics.

=> v = u + at

=> 0 = 5 + (-10) t

=> 5 = 10t

=> 0.5 seconds.

Therefore,

Height attend by the stone = 1.25 m

Time take to reach = 0.5 seconds.