Is this projectile motion situation possible?

[McMaster27]

Homework Statement

An archer fires an arrow which travels in the air for 2.5 seconds and strikes a target 75 m away. Assuming that the target is at the same height that arrow is launched from, what was the initial velocity of the arrow if it was shot at an angle of 25° above horizontal?

Relevant Equations

Vx = dx/t

Hi everyone,
I have created a question which I thought would have a single simple solution, but have noticed there are two possible answers. This makes me think that the question's scenario is impossible with the numbers I made up.

I think we all can agree that the horizontal component to velocity is 30 m/s (this is solved by v = d/t = 75m/2.5s). The discrepancy comes about though if you then try to solve for the initial vertical velocity. Two different methods provide two slightly different answers. A vector diagram could be set up, and use 30m/s * tan 25 = viy which comes out to 13.99 m/s [up]. However, if you use the kinematic equation d = viy*t + 1/2at^2, then you get a slightly different answer of 12.25 m/s [up].

Can anyone identify why this discrepancy exists especially when I am using exact values and not rounding in those calculations?

 

[PeroK]

I suspect your problem is over specified. Which means there is no solution for the data you have specified.

 

[McMaster27]

I suspect your problem is over specified. Which means there is no solution for the data you have specified.

Is it possible to find the right variables that would make the problem come out with one exact answer? The issue probably lies with the angle I provided, but certainly there must be SOME specific angle that would allow a projectile to travel 75m in 2.5 seconds. How could I find this?

 

[erobz]

Is it possible to find the right variables that would make the problem come out with one exact answer? The issue probably lies with the angle I provided, but certainly there must be SOME specific angle that would allow a projectile to travel 75m in 2.5 seconds. How could I find this?

Without specifying an angle, solve for $v$ with both equations. The equate them to solve for the angle $\theta$. Then solve for velocity once the angle is determined.

Over specified means you must clear the way for something to adjust.

 

[McMaster27]

Without specifying an angle, solve for $v$ with both equations. The equate them to solve for the angle $\theta$. Then solve for velocity once the angle is determined.

Over specified means you must clear the way for something to adjust.

Thank you, I've done that and realized the scenario is possible with an angle of 22 degrees now.

The hand slapping the forehead emoji is what I look like right now! lol

 

[kuruman]

Is it possible to find the right variables that would make the problem come out with one exact answer? The issue probably lies with the angle I provided, but certainly there must be SOME specific angle that would allow a projectile to travel 75m in 2.5 seconds. How could I find this?

Yes. Once you have fixed the range $R$ and the time of flight $t_{\!f}$ to anything you want, the projection angle $\theta$ is fixed. In terms of the given quantities $$\tan\theta=\frac{v_{0y}}{v_{0x}}=\frac{\frac{1}{2}g~t_{\!f}}{\frac{R}{t_f}}=\frac{g~t_{\!f}^2}{2R}.$$

 

[haruspex]

To check for overspecification, count the degrees of freedom,
If you were told the launch speed, the launch angle and the range, everything else (relative to the launch point) would be determined. So the solver only needs three facts, but you provided four.