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1) This is not a homework problem, just a question I came across and can't solve adequately.
2) Sorry if this is wrong area to post it.
What angle would be best to hold your head at relative to ground to maximize the distance you can spit.
I drew a diagram, horizontal axis was distance in the x direction, vertical axis was distance in y direction. I drew a vector for velocity, V0 at an angle \theta relative to ground. vx and vy represent the x and y components of this initial velocity vector. The initial velocity starts from a height of h.
I then divided it into 2 equations, the acceleration in the y and x directions.
The Y direction:
ay = -9.8 (m/s2)
integrate twice to obtain position function:
py(t) = -4.9t2 + vyt + h
Next, I found the time, t1, that the projectile (saying spit sounds weirder) would reach ground, at y = 0.
py(t1) = 0
using the quadratic equation I obtained 2 results, 1 of which had to be negative. So the other one is the time we want.
t1 = Vy/9.8 + (1/9.8)(\sqrt{v<sub>y</sub><sup>2</sup> + 19.6h})<br /> <br /> OK, so at this time, we will have y = 0 and x = D, the farthest distance the projectile will reach.<br /> <br /> The x equation:<br /> a<sub>x</sub> = 0 --> v<sub>x</sub>(t) = v<sub>x</sub> --> p<sub>x</sub>(t) = v<sub>x</sub>t<br /> <br /> replace t with the value found for t1, because that will give us an equation for D, the maximum distance.<br /> v<sub>x</sub>(t1) = D = \frac{v<sub>x</sub>v<sub>y</sub>}{9.8} + (1/9.8)(v<sub>x</sub>\sqrt{v<sub>y</sub><sup>2</sup> + 19.6h)
2) Sorry if this is wrong area to post it.
Homework Statement
What angle would be best to hold your head at relative to ground to maximize the distance you can spit.
The Attempt at a Solution
I drew a diagram, horizontal axis was distance in the x direction, vertical axis was distance in y direction. I drew a vector for velocity, V0 at an angle \theta relative to ground. vx and vy represent the x and y components of this initial velocity vector. The initial velocity starts from a height of h.
I then divided it into 2 equations, the acceleration in the y and x directions.
The Y direction:
ay = -9.8 (m/s2)
integrate twice to obtain position function:
py(t) = -4.9t2 + vyt + h
Next, I found the time, t1, that the projectile (saying spit sounds weirder) would reach ground, at y = 0.
py(t1) = 0
using the quadratic equation I obtained 2 results, 1 of which had to be negative. So the other one is the time we want.
t1 = Vy/9.8 + (1/9.8)(\sqrt{v<sub>y</sub><sup>2</sup> + 19.6h})<br /> <br /> OK, so at this time, we will have y = 0 and x = D, the farthest distance the projectile will reach.<br /> <br /> The x equation:<br /> a<sub>x</sub> = 0 --> v<sub>x</sub>(t) = v<sub>x</sub> --> p<sub>x</sub>(t) = v<sub>x</sub>t<br /> <br /> replace t with the value found for t1, because that will give us an equation for D, the maximum distance.<br /> v<sub>x</sub>(t1) = D = \frac{v<sub>x</sub>v<sub>y</sub>}{9.8} + (1/9.8)(v<sub>x</sub>\sqrt{v<sub>y</sub><sup>2</sup> + 19.6h)