Question

If the range of a projectile's trajectory is ten times larger than the height of the trajectory, then what was the angle of launch with respect to the horizontal? (Assume a flat and horizontal landscape.)

Answer 1

Answer:

Angle of projection from the horizontal will be $21.80^{\circ}$

Explanation:

We have given that range of the projectile is 10 times the height of the projectile

Let the projectile is projected with velocity u at an angle $\Theta$

Range of the projectile is equal to $R=\frac{u^2sin2\Theta }{g}$

And height of the projectile is equal to $h=\frac{u^2sin^2\Theta }{2g}$

Now according to question range is 10 times of height

So $\frac{u^2sin2\Theta }{g}=10\times \frac{u^2sin^2\Theta }{2g}$

$sin2\Theta =5sin^2\Theta$

$2sin\Theta cos\Theta =5sin^2\Theta$

$tan\Theta =\frac{2}{5}=0.4$

$\Theta =tan^{-1}0.4=21.80^{\circ}$

So angle of projection from the horizontal will be $21.80^{\circ}$