Arrow being fired to centre of target

[hello478]

Homework Statement

part b and part c of the question

Relevant Equations

suvat equations

i solved it like this...
s = ut + 1/2 at^2
t= 1.08 (from part a)
u= 65 sin4.30
a= 9.81? or -9.81
the answer said -9.81
why? wouldn't acceleration change from -9.81 to +9.81 because it moves up then down???
its soo confusing...

 

[Mark44]

Homework Statement: part b and part c of the question
Relevant Equations: suvat equations

i solved it like this...
s = ut + 1/2 at^2
t= 1.08 (from part a)
u= 65 sin4.30
a= 9.81? or -9.81
the answer said -9.81
why? wouldn't acceleration change from -9.81 to +9.81 because it moves up then down???

No, the acceleration wouldn't change unless gravity suddenly reversed.

its soo confusing...

View attachment 342395

 

[hello478]

No, the acceleration wouldn't change unless gravity suddenly reversed.

ok... so why is acceleration -9.81?

i fixed the picture now...

 

[Mark44]

ok... so why is acceleration -9.81?

Because they're assuming that the "up" direction is positive, and gravity is acting downward.

 

[hello478]

Because they're assuming that the "up" direction is positive, and gravity is acting downward.

is it in the question? i cant find it...

 

[Mark44]

is it in the question? i cant find it...

It's not explicitly given. What they tell you is that the speed (i.e., $|\vec v|$) of the arrow is 65 m/sec, at an angle of 4.3° above horizontal. The arrow's velocity vector can be decomposed into a vertical component and a horizontal component.

The usual approach for problems of this sort is to treat upward velocities as positive, with gravity acting downward (so g = -9.81 m/sec^2). During the flight, the arrow's vertical component of velocity will start off positive, slowly decrease to zero at its high point, and then become negative as it continues on to the target.

 

[hello478]

ok thank you, i got it now