A projectile’s horizontal range on level ground is R= [(Vo)^2 sin2θ] / g. At what launch angle or angles will the projectile land at half of

0 thoughts on “A projectile’s horizontal range on level ground is R= [(Vo)^2 sin2θ] / g. At what launch angle or angles will the projectile land at half of”

1. Answer:

   15° and 75°

   Explanation:

   Given:

   The range of a projectile motion is given as:

   [tex]R=\frac{V_0^2sin2\theta}{g}[/tex]

   Where,

   [tex]V_0\to initial\ velocity\\\\\theta\to angle\ of\ projection\\\\g\to\ acceleration\ due\ to\ gravity=9.8\ m/s^2[/tex]

   Maximum possible range is when [tex]\sin 2\theta=1[/tex]

   So, maximum range possible is, [tex]R_{max}=\frac{V_0^2}{g}[/tex]

   Now, as per question:

   [tex]R=\frac{R_{max}}{2}\\\\\frac{V_0^2\sin 2\theta}{g}=\frac{V_0^2}{2g}\\\\\sin 2\theta=\frac{1}{2}\\\\\sin 2\theta=\sin 30\ or\ sin 2\theta=\sin 150\\\\2\theta=30\ or\ 2\theta=150\\\\\theta=\frac{30}{2}=15\ or\ \theta=\frac{150}{2}=75[/tex]

   Therefore, the launch angles are 15 degree and 75 degree.

   Reply