Question

A projectile is fired in such a way that its horizontal range is equal to 12.5 times its maximum height. What is the angle of projection?

Answer 1

Use the formulas!!

That's it!! :D

Answer 2

Im trying to figure this out. I'm using pretty common notation. Vyf, Vxf, Vyi, Vxi, R, and h

Answer 3

I'm trying to use the formulas but idk how to connect the dots

Answer 4

h = -gt^2/2 + Vo sin(a) t

d = t cos(a)

Answer 5

forgot a Vo in my d

Answer 6

the vertex of the parabola would be half the horizontal ... might want to keep that in the back of the picture

Answer 7

R = u^2 sin 2theta/g
H = u^2 sin^2 theta/2g

R = 12.5 * H

Now find theta!! :D

Understood? :)

Answer 8

http://en.wikipedia.org/wiki/Projectile_motion

Answer 9

\[h=-\frac12gt^2+V_osin(\alpha)~t\]

\[d=V_ocos(\alpha)~t=12.5h\]when:\[h=\frac{V_o}{12.5}cos(\alpha)~t\]

Answer 10

\[h=-\frac12gt^2+V_osin(\alpha)~t\]
\[h'=-gt+V_ocos(\alpha)=0~:~t=\frac{V_ocos(\alpha)}{g}\]
\[h_{max}=-\frac12g\left(\frac{V_ocos(\alpha)}{g}\right)^2+V_osin(\alpha)~\frac{V_ocos(\alpha)}{g}\]
\[d_{max}=V_ocos(\alpha)~t,~when~h=0\]
\[0=t\left(-\frac12gt+V_osin(\alpha)\right)\]
\[t=0, ~t=\frac{2~V_osin(\alpha)}{g}\]
\[d_{max}=V_ocos(\alpha)~\frac{2~V_osin(\alpha)}{g}=12.5~h_{max}\]

Answer 11

\[V_ocos(\alpha)~\frac{2~V_osin(\alpha)}{g}=-6.25g\left(\frac{V_ocos(\alpha)}{g}\right)^2+12.5V_osin(\alpha)~\frac{V_ocos(\alpha)}{g}\]

\[V_ocos(\alpha)~\frac{2~V_osin(\alpha)}{g}-12.5V_osin(\alpha)~\frac{V_ocos(\alpha)}{g}=-6.25g\left(\frac{V_ocos(\alpha)}{g}\right)^2\]

\[-\frac{10.5}{g}(V_o)sin(\alpha)cos(\alpha)=-6.25g\left(\frac{V_ocos(\alpha)}{g}\right)^2\]

\[-{10.5}(V_o)sin(\alpha)cos(\alpha)=-6.25g^2\left(\frac{V_ocos(\alpha)}{g}\right)^2\]

\[{10.5}(V_o)sin(\alpha)cos(\alpha)=6.25(V_o)^2 cos^2(\alpha)\]

\[{10.5}sin(\alpha)cos(\alpha)=6.25V_o~cos^2(\alpha)\]

\[{10.5}sin(\alpha)=6.25V_o~cos(\alpha)\]

Answer 12

the Vo goes away, i missed it on the left above ....

Answer 13

\[{10.5}sin(\alpha)=6.25~cos(\alpha)\]
\[\frac{10.5}6.25tan(\alpha)=1\]
etc..

Answer 14

the latex went wonky on me ..

(10.5/6.25) tan(a)=1

tan(a)= 6.25/10.5

a= arctan(6.25/10.5)

maybe