Question

a stone is thrown in a vertically upward direction with a velocity of 5 metre per second if the acceleration of the stone during its motion is 10 metre per second in the downward Direction What will be the height attained by stone and how much time will it take to reach there I don't need and useless answer​

Answer 1

Given :

The upward velocity of the stone = 5 m/s

Acceleration of the stone = 10 m/s (Downward direction)

To Find :

• The maximum height reached by the stone

• Time taken to reach maximum height

Solution :

• From equations of motion

         $v^{2} -u^{2} =2as$

• Maximum height attained by the stone

        $Height =s = \frac{v^{2} - u^{2} }{2a}$

                        $s = \frac{0-25}{2\times(-10)}$

                        $s = 1.25m$

    The maximum height attained by the stone is 1.25m

• Time taken to reach maximum height

            $v = u+at$

            $t = \frac{v-u}{a}$

            $t = \frac{-5}{-10}$

            $t = 0.5sec$

    The time taken by the particle to reach maximum height is 0.5sec

Answer 2

Answer:

Given :

The upward velocity of the stone = 5 m/s

Acceleration of the stone = 10 m/s (Downward direction)

To Find :

The maximum height reached by the stone

Time taken to reach maximum height

Solution :

From equations of motion

v^{2} -u^{2} =2asv

2

−u

2

=2as

Maximum height attained by the stone

Height =s = \frac{v^{2} - u^{2} }{2a}Height=s=

2a

v

2

−u

2

s = \frac{0-25}{2\times(-10)}s=

2×(−10)

0−25

s = 1.25ms=1.25m

The maximum height attained by the stone is 1.25m

Time taken to reach maximum height

v = u+atv=u+at

t = \frac{v-u}{a}t=

a

v−u

t = \frac{-5}{-10}t=

−10

−5

t = 0.5sect=0.5sec

The time taken by the particle to reach maximum height is 0.5sec

Explanation:

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