Question

A stone is thrown in a vertically upward direction with a velocity of 5 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:

maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Explanation:

the acceleration of the stone during its motion is 10 m 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? Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Answer 2

Given :

• Initial velocity, u = 5 m/s
• Acceleration, a = - 10 m/s²  
• Final velocity, v = 0

To Find :

• Time taken, t = ??
• Distance covered, s = ??

Formula used :

1st equation of motion,

${\underline { \boxed { \bf { \pink{v = u + at}}}}} \: \green \bigstar$

3rd equation of motion,

${\underline { \boxed { \bf { \red{v ^{2} - u ^{2} = 2 s}}}}} \: \blue \bigstar$

Solution :

• Putting all the values, we get

$: \implies \tt \:v = u + at$

$: \implies \tt \: 0 = - 5 + (- 10) × t$

$: \implies \tt - 5 = - 10 t$

$: \implies \tt \: \frac{5}{10} = t$

$: \implies \tt \: t= \frac{1}{2}$

$: \implies \tt \: t = 0.5 seconds$

Hence, the time taken by stone to reach there is 0.5 seconds.

Now,

using

$: \implies \tt \: v ^{2} - u^{2} = 2 \: as$

$: \implies \tt(0) ^{2} - (5)^{2} = 2 × (- 10) × s$

$: \implies \tt - 25 = - 20 × s$

$: \implies \tt \frac{- 25 }{- 20} = s$

$: \implies \tt s = 1.25$

Hence, the height attained by stone is 0.5 seconds.