How do i work out this physics question which is. A stone is dropped from a height of 52m above the ground?

A stone is dropped from a height of 52.0m above the ground


a) how long does it take to reach the ground?


b) what is its velocity just before it hits the ground?


c) how far does it fall in the last second of its motion?


Please answer them and show working :) and i will give you 10 points!!!How do i work out this physics question which is. A stone is dropped from a height of 52m above the ground?
Ok here goes..


s=52...(displacement)


s=ut+1/2gt^2...since it is acting under the effect of gravity..a=g..


since it is 'dropped'..it has no initial velocity..so u=0..


so 52=0+1/2(9.8)t^2..


52=4.9t^2...


t=sqrt(52/4.9)...


v=u+at...v=u+gt


u=0,


so v=gt...substitute the value for t and g





in this case we assume that the stone comes down with an intial velocity and its final velocity becomes 0 on reaching the ground...


so v=0..u=v of the previous case..wat u calculated to be gt..


v^2=u^2+2gs...v=0,


u^2=-2gs...to find u substitute values for g and s..(use -9.8 for g)





ok u found t in the first case....


substitute that value in the eqn s=ut+1/2gt^2...substitute 0 for u and find s using the known values of g and t...Hope i helped..How do i work out this physics question which is. A stone is dropped from a height of 52m above the ground?
s=52m


g=9.8m/s2





s=g*t*t/2 as initial v=0


so t=3.26





v*v=2gs


v=31.92





s2-s1=52-g*(t-1)*(t-1)/2


s2-s1=26.98
I'll help, but promise a best answer.





So .