Yes, your calculations are correct.

[petuniac]

you are playing darts and are wondering how far the dart is dropping on its journey to the target which is 4 m away. The dart leaves your hand at 7 m/s and is accelerating at a rate of -0.3 m/s^2.

work so far...

find final velocity is x-direction from vf^2 = vi^2 + 2ad

vf = 6.8 m/s when dart strikes target

find time it takes to hit target from

vf = vi + at

t = 0.67 s

now, find displacement in y-direction from

d = vit + 1/2at^2
d = -2.2 m

is this correct?

thanks

 

[OlderDan]

The method looks good, the time does not.

 

[QuantumCrash]

It seems like linear kinematics! Really, I would have thought something like this would be projectile motion.

 

[OlderDan]

It seems like linear kinematics! Really, I would have thought something like this would be projectile motion.

It is projectile motion, but with a retarding force. The question is poorly worded because it is not specific about the direction of the given acceleration, or the initial velocity. It clearly does not include the gravitational acceleration in what is given, so the asumption that it is a horizontal acceleration of -0.3m/s^2 is most likely what they expected. Since no initial angle was specified, and they are asking for "dropping" distance, a horizontal initial velocity was likely intended.

 

[petuniac]

thanks for the replies... please see my comments below

1) why does the time not look correct? I re-did the calculation with the same result

vf = vi + at
6.8 = 7 + (-0.3)t
t = (6.8-7)/-0.3
t = 0.67 s

2) this problem was given as part of linear kinematics...

3) so am I doing this correctly??

thanks!

 

[OlderDan]

thanks for the replies... please see my comments below

1) why does the time not look correct? I re-did the calculation with the same result

vf = vi + at
6.8 = 7 + (-0.3)t
t = (6.8-7)/-0.3
t = 0.67 s

2) this problem was given as part of linear kinematics...

3) so am I doing this correctly??

thanks!

You rounded off too much computing the 6.8m/s. The difference between the two velocities is too small to get an accurate time unless you keep more digits. You could of course argue that the problem did not give the initial velocity very accurately, but when they say 7 they really mean 7.00000000000000000000000000000000. You are doing it correctly for the assumptions I outlined earlier. It's really a poorly worded problem. I think you did what was intended.