Hey, i feel kinda embarassed asking this because it looks so simple.

a mass of 300kg which is positioned 1000m above the ground is allowed to fall to the ground. what will be its velocity on impact?

Use the basic velocity formula. Initial velocity is zero; solve for final velocity. Your acceleration is gravity.

so i should use

v^2 = u^2 + 2as

(v- final and u- initial)??

i did

v^2= 0+ 2(9.81)(1000)
= 19620

and then i find the square root of that which is 140 m/s

is that correct?

Initial PE= final KE
mgh= 1/2 mv^2

V^2= 2gh= 2*1000*9.8
V= 140m/s

Yes, you have correctly used the basic velocity formula to find the velocity of the mass on impact. The formula you used is v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration (gravity in this case), and s is the displacement (the height the mass falls, which is 1000m in this case).

By plugging in the values, you correctly found that v^2 = 0 + 2(9.81)(1000) = 19620. To find the final velocity, you take the square root of this value, giving you v = √19620 ≈ 140 m/s.

Another way to approach this problem is by using the principle of conservation of energy. The initial potential energy (PE) of the mass when it is positioned at a height of 1000m is equal to the final kinetic energy (KE) of the mass when it hits the ground.

The equation for potential energy is PE = mgh, where m is the mass (300kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (1000m). So, the initial potential energy is PE = (300)(9.8)(1000) = 2,940,000 J.

The equation for kinetic energy is KE = 1/2 mv^2, where v is the velocity of the mass on impact. Since we want to solve for v, we can set KE equal to the initial potential energy:

KE = 1/2 mv^2 = 2,940,000 J

By rearranging the equation, we can solve for v:

v^2 = 2KE/m = 2(2,940,000)/(300) ≈ 19620
v = √19620 ≈ 140 m/s

So both methods give the same result of v ≈ 140 m/s, which is the velocity of the mass on impact.