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i)
v=0
u= ?
a= -10
s=40+5 = 45
using v^2 = u^ 2 + 2*a*s
u^2= 900
u= 30
therefore speed of projection = 30 m/s
ii)
s=5
u=0
a=10
using s= ut + 0.5*a*t^2
5= 0+ 0.5*10*t
t=1
time above top of cliff = 1*2 = 2s
this is because the signal turns n falls back. so it takes as much time to fall back (same 5m) as it took to go up the 5 m above the cliff.
iii)
s= ?
u= 0
t= root 17
find the height reached by the signal (max height of signal)
s= ut+ 0.5*a*t^2
s= 0+ 0.5*10*( root 17/2) ^2
s= 21.25
time is divided by 2 then squared, ths is because root 17 is the total time tht the signal was abovet the top of tht cliff. means going up + coming down time
finding the speed:
use v^2 =u^2 +2as
v= 0
s=21.25 + 40 =61.25
a= -10
0= u^2 -2(10)(61.25)
u= root 1225 = 32 m/s
