 So, this question is, a body is thrown vertically upward with velocity u, the greatest height h to which it will rise, so we'll basically find out, you have to find out how much height the body go to. So, in such questions, it's very good practice to draw a diagram, so let's say this is the ground level and the body is being thrown up and it goes up to let's say this height. So, initially the body was here, initially the body was here and then finally, let's say the body reached this point, so let's say this is the height h which it attains, let's say this is my datum line and this is my positive y direction. So, if the body is going up, so in this case, displacement is given as positive h, so since it is going from ground to sky, towards the sky, so positive h, now so and it's also given that it is thrown by initial velocity u, so it will attain the maximum height, what is the catch here? So, maximum height, max height will be attained where, what is the condition of maximum height? The maximum height will be where the final velocity, final velocity v is equal to 0, v equals to 0. Now, so we know that v equals to 0, u is given as u and we have to find out h, now let's write down all the three equations of motion, so first equation of motion was v is equal to u plus a t, 2 is s is equal to u t plus r a t square and third is v square is equal to u square plus 2 a s, so in these three equations, the assumptions are v, u, s and a, all are in the same direction, same direction, if one of them is in the reverse direction or opposite direction, the sign of that quantity will be changing, so in this case, u was towards the sky, so positive, h is towards the sky, so positive, v is 0, so doesn't matter, what is acceleration in this case? Acceleration is nothing but acceleration due to gravity which is g and is downward, so the sign which we will be using while calculating will be negative g, now the value of g is 9.8 meter per second square which is roughly 10 meter per second square, though we will not be using the value here, but just for the sake of information, now so obviously there is no mention of t, so we can very well use this equation, third one, third one, why? Because there is mention of a we know s, we know u, so basically we have to find out s, now so using the third equation, so using equation 3, we get v square is equal to u square minus 2 gs, 5 because g is having a direction which is opposite to all v, u and s, so hence negative and we have considered positive direction towards the sky, so now let's deploy the value, so 0 meter per second is equal to u square, so since there is everything is invariable, so we need not mention the unit, so this is 0, u square minus 2 g into h, h is the height obtained or height to which the body goes, this implies I can write 2 gh is equal to u square or h will be equal to, this implies h will be equal to u square upon 2g, so this is the final answer which happens to be option b.