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Phosu:
yes it does land below the starting point...

A ball is projected vertically upwards with a speed u m s?1 from a point A which is 1.5 m above the ground. The ball moves freely under gravity until it reaches the ground. The greatest height attained by the ball is 25.6 m above A.
The ball reaches the ground T seconds after it has been projected from A.
(a) Show that u = 22.4
(b) Find, to 2 decimal places, the value of T.

nid404:
a)
v2=u2+2as
velocity and max height=0
0=u2- 2gX25.6  ( - because ball decelerates when it's moving upwards)
u2=2gX25.6
u=22.4m/s

b) v=u+at for the first part(upward motion)
0=22.4-10t
t=2.24s to reach max height

comin down (25.6+1.5)m
s=1/2 Xat2
27.1=1/2X10 t2
t=2.32s
total time= 2.24+2.32=2.56s

simba:
i hv a question :   


A load of mass 50kg stands on the horizontal floor of a goods lift which is accelerating vertically downwards at 2 m/s square. Calculate the magnitude of the contact force exerted on the load by the floor.

astarmathsandphysics:
50(9.8-2)=390N

simba:
the answer is 400, nt 390 .......i saw it bt i dnt knw how it is done

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