Mechanics that shape our world!

Sorry guys my mistake I accidentally copied the answer too. <-_ -> sob sob

h=v(initial)*t + 1/2 gt^2 .......... ...........v(initial)=0 .................h= 1/2 gt^2 .............v=x/t ........... t=x/v .............t^2=(x/v)^2. .......................... x=v sqrt(2h/g) HINT: FIST LAW (H=V(I)T+.5 GT^2) SOLVED FOR VERTICAL COMPONENT but second one (v=x/t) solved for horizontal component THANKS! by yuri

h=0.5gt^2.x=vt.

That is nothing but Range=velocity ×time

we should just observe the motion in the horizontal direction to find horizontal range and the motion in the horizontal direction is uniform motion so we can use distance=speed x time and for finding the time for which the motion is executed, we can see the motion in the vertical direction initially the velocity in the vertical direction is 0 and it has to cover a displacement 'h' the motion in the vertical direction is uniformly accelerated motion so we can use the equation s=ut+(at^2)/2 where, s is the displacement(here it's h) , u is the initiial velocity (here it is0) , t is the time taken (we have to find this) and a is the acceleration(here it is 'g' as only one force , the force of gravity is acting on the object so acceleration = force / mass =mg/m = g) so we get the time taken for the motion from here and then we can calculate the distance travelled in the horizontal direction(range) i hope this helps you:)