Question

A ball is thrown in vertically upward direction with a velocity of 10 m/s. if the
acceleration of the stone during its motion is 10 m/s2 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

Answer:

given : u = 10m/sec, g = -10m/sec², v = 0

to find, s and t

v² - u² = 2gs

0² - 10² = 2×(-10)×s

-100 = -20 s

s = -100/-20

= 5 m

v = u + gt

0 = 10 + (-10) t

0 = 10 - 10t

0 = 10 (1 - t)

1 - t = 0/10

1 - t = 0

t = 1 sec

OR

s = ut + 1/2gt²

5 = 10t + 1/2(-10)t²

5 = 10t - 5t²

5t² - 10t + 5 = 0

t² - 2t + 1 = 0

t² - t - t + 1 = 0

=> t (t - 1) - 1 (t - 1) = 0

=> (t - 1)(t - 1) = 0

t = 1 sec

Answer 2

$\large \underline{\underline{\mathfrak{Answer :}}}$

• Time interval is 1 second
• Height attained by ball is 5 m

$\mathfrak {\underline{\underline{\mathfrak{Step - By - Step - Explanation :}}}}$

GiveN :

• A ball is thrown vertically upwards with velocity of 10 m/s.
• Acceleration is 10 m/s²

To FinD :

• Height attained by ball.
• Time taken to reach height.

SolutioN :

• We are given that a ball is thrown vertically upwards with a velocity of 10 m/s. It means initial velocity is 10 m/s

• And the acceleration of the ball will be - 10 m/s² because acceleration is applied in opposite direction of motion.

• Ball will stop after reaching height. So, final velocity will be 0 m/s.

By above data we have :

• Initial velocity (u) = 10 m/s
• Acceleration (a) = - 10 m/s²
• Final velocity (v) = 0 m/s

___________________________

Use 1st equation of motion for calculating time interval.

$\longrightarrow \sf{v \: = \: u \: + \: at} \\ \\ \longrightarrow \sf{0 \: = \: 10 \: + \: (-10) \: \times \: t} \\ \\ \longrightarrow \sf{0 \: - \: 10 \: = \: -10t} \\ \\ \longrightarrow \sf{-10t \: = \: -10} \\ \\ \longrightarrow \sf{t \: = \: \dfrac{-10}{-10}} \\ \\ \longrightarrow \sf{t \: = \: 1} \\ \\ \underline {\boxed{\sf{Time \: = \: 1 \: s}}}$

Time taken is 1 second.

_______________________________

For calculating distance, use 3rd equation of motion :

$\longrightarrow \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \longrightarrow \sf{0^2 \: - \: -10^2 \: = \: 2 \: \times \: - 10 \: \times \: s} \\ \\ \longrightarrow \sf{-100 \: = \: -20s} \\ \\ \longrightarrow \sf{s \: = \: \dfrac{100}{20}} \\ \\ \longrightarrow \sf{s \: = \: 5} \\ \\ \underline{\boxed{\sf{Height \: = \: 5 \: m}}}$

Height up-to which ball reached is 5 m