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M1 Kinematics help

10. A stone is dropped from the top of a tower. One second later another stone is thrown vertically downwards from the same point with a velocity of 14m/s. If they hit the ground together, find the height of the tower.

Ok so I have:

stone1:
s=s
u=u
v=v
a=-9.8
t=t

stone2:
s=s
u=14
v=v
a=-9.8
t=(t+1)

how on earth do I do this? Is my information even correct? Many thanks :smile:
Reply 1
(I'll be taking down as positive so acceleration will be positive in this case)

Stone 1: u = 0ms^-1, t = T, a = 9.8ms^-2
Stone 2: u = 14ms^-1, t = T-1, a = 9.8ms^-2

Use s = ut + 1/2 at^2 and equate both equations :smile:
Reply 2
Original post by CTArsenal
(I'll be taking down as positive so acceleration will be positive in this case)

Stone 1: u = 0ms^-1, t = T, a = 9.8ms^-2
Stone 2: u = 14ms^-1, t = T-1, a = 9.8ms^-2

Use s = ut + 1/2 at^2 and equate both equations :smile:

why on earth do we minus 1 of the initial velocity?
Reply 3
Original post by Mr Tall
why on earth do we minus 1 of the initial velocity?


14ms^-1 is the same as 14m/s :tongue:

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