Question

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

Given that:

Initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms-2 in the direction opposite to the trajectory of the stone = -10 ms-2

Find out:

The height attained by the stone and how much time will it take to reach there

Formula:

As per the third motion equation, v2 – u2 = 2as

Therefore, the distance traveled by the stone (s) = (02 – 52)/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

plz mark me as brainliest

Answer 2

Appropriate Question :

• A stone is thrown in a vertically upward direction with a velocity of 5m-1. If the acceleration of the stone during its motion is 10ms-2? in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer :

• Height = 1.25 , time = 0.5sec

Given :

• A stone is thrown in a vertically upward direction with a velocity of 5ms-¹
• Acceleration of the stone during its motion is 10ms-²

To find :

• in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution :

Given,

• Initial velocity (u) = 5ms-1
• Final velocity (v) = 0

Here , Given, Initial velocity is upward direction and acceleration is in the downward then acceleration will become negative sign,

• a = 10ms-2

• Let the height be h
• Let the time be t

Finding the height

We know that third equation of motion formula :

• v² = u² + 2as

Where,

• v is final velocity
• u is Initial velocity
• a is acceleration
• s is height

➠ v² = u² + 2as

➠ (0)² = (5)² + 2 × -10 × h

➠ 0 = 25 + 20 × h

➠ 20h = 25

➠ h = 1.25m

Now Finding the time by using the first equation of motion formula :

• v = u + at

where,

• v is final velocity
• u is Initial velocity
• a is Acceleration
• t is time

➠ 0 = 5 + -10 × t

➠ 10t = 5

➠ t = 5/10

➠ t = 0.5 sec

Hence , Height = 1.25 , time = 0.5sec

More Explanation :

• v = u + at
• s = ut + 1/2 at²
• v² = u² + 2as