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Need help for Space Q again (1 Viewer)

nightweaver066

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a) -40 = -9.8t
t = 4.08s

b) s = 30 x t
= 122.15m

c) s = ut + 1/2 at^2
-25 = 1/2 (-9.8)t^2
t = sqrt(50/9.8)
= 2.26s

d) s = ut + 1/2 at^2
s = 1/2(-9.8)(2)^2
= -19.6m

40 - 19.6 = 20.4m
 

_deloso

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a) s=ut+1/2at^2 where u is initial vertical velocity which equals zero because it is thrown horizontally
40=1/2(9.81)t^2
t^2=40/4.905
t= 2.85568...
t=2.86

b) R=vt
R=30x2.86
R=85.67m

c) 40-25 = 15
therefore, distance is 15m
s=ut+1/2at^2
15=1/2(9.81)t^2
t^2=15/4.905
t=1.7487...
t=1.75s

d)s=ut+1/2at^2
s=1/2(9.81)(2)^2
s=19.62
distance traveled after 2 sec is 19.62 metres...
therefore, height above ground: 40-19.62 = 20.38 metres above ground
 

cheese_cheese

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After the HSC no one will care about this crap. Good luck for the HSC though :)
 

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