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paperdoll
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Homework Statement
The bridge is 9.5 m above the water and using his watch, Andrew measured the time it took for a stone, thrown by David to return to the height of the bridge to be .5 seconds.
a. what is the vertical velocity with which the stone left David's hand?
b. how high did the stone go over the water?
c. what was the magnutude of the stones velocity in the horizontal direction if it left david's hand at an angle of 38 degrees?
d. what distance from the base of the bridge did the stone travel?
Homework Equations
s=ut+1/2at^2
The Attempt at a Solution
The thing I am confused about is whether to put the acceleration negative or positive.
I let Vs=0
and then
0=ut+1/2at^2
0=0.5u-9.8*1/2*(0.5)^2
u=2.45 m s^-1
for part b) I did
At max height v=0
0=2.45^2+2*-9.8*s
s=0.306
so max height is 9.5+0.306?= 9.806 m
c)
tan 38=2.45/uH
uH=3.1358 ms^-1
d) sH=uH*t
3.1358*t
v^2=2.45^2-2*9.8*-9.5
...
and then I get a negative answer that cannot be square rooted :\
Is what I'm doing right? I'm not sure on the direction of acceleration :\
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