Projectile motion with 15° angle and 3.5s flight time

[squall325]

1. Angle = 15, Time(s) = 3.5, Horizontal Distance(m) = ?, Vertical Distance(m) = ?, Initial Velocity(m/s) = ?, Final Velocity(m/s) = ?



2. Our teachers told us to use y = vi(sine(theta)(t) - (gt^2)/2 - but we can't get an answer with only 2 givens.



3. Cant get an answer using the given equation. Hope you guys could help.

 

[Matthaeus_]

Indeed, you can't solve for so many unknowns.

 

[Dick]

Indeed, you can't solve for so many unknowns.

Sure you can, if you assume 'time' is total time of flight and the surface is horizontal (etc etc). y=0 at t=0 and y=0 at t=3.5 sec. Use that to solve for vi and vertical distance. Now use x=vi*cos(theta)*t to do the horizontal part.

 

[robsmith82]

Maybe you are to assume the object is to be thrown from ground level? Then this is solvable.

 

[robsmith82]

Maybe you are to assume the object is to be thrown from ground level? Then this is solvable.

 

[Matthaeus_]

Ah, well, putting $$y_0 = 0$$ then you can solve:

$$\displaystyle \left \{ \begin{array}{ll}<br /> y &= (v_0 \sin 15^{\circ})t - \frac{1}{2} gt^2\\<br /> v_y &= (v_0 \sin 15^{\circ}) - gt\\<br /> x &= (v_0 \cos 15^{\circ})t<br /> \end{array}$$

 

[andrevdh]

You can solve the parabolic equation

$$y = x \tan(\theta _o) - \frac{gx^2}{2v_o \cos^2(\theta _o)}$$

for the initial velocity $$v_o$$ - it is the only unknown in the equation if you use the other equations to subs for x and y.

But the problem do not require the initial velocity to be confined by any of the given values. This means that even if the projectile is resting on the launching pad at fifteen degrees and time marches on for 3.5 seconds the conditions are still met! So one can shoot the projectile at any initial speed and the situation will still be within limits of the given data.