Question

A stone is thrown vertically upwards with a velocity 40 m/s . At what height k.E and p.E will be equal

Answer 1

Answer : -

Given that,

Initial velocity (u) = 40 m/s

Final velocity (v) = 0 m/s

Assumption : -

Let's assume, acceleration due to gravity = - 10 m/s^2 [for our ease of calculation].

We have taken it negatively because the stone is goin' in a upward direction and thus, it's opposing the gravitational force.

By using third equation of motion,

2as = v² - u²

2(-10)s = (0)² - (40)²

2(-10)s = 0 - 1600

- 20s = - 1600

$\cancel{-}20s = \cancel{-} 1600$

$s = \frac{1600}{20}$

$s = \frac{160 \cancel{0}}{2 \cancel{0}}$

$s = \frac{160}{2}$

s = 80 m

Therefore, the distance covered by the stone is 80 m.

But at half distance, the kinetic energy (K.E.) and potential energy (P.E.) are same.

That is, distance = $\frac{Total\:distance}{2}$

=> Distance = $\frac{80\:m}{2}$

=> Distance = 40 m

Therefore, at 40 m the kinetic energy (K.E.) and potential energy (P.E.) are same.

$\huge{ \boxed{ \pink{distance = 40 \: m}}}$

Answer 2

Answer:

distance is 40m

Explanation:

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