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maxonik [38]
3 years ago
9

If the range of a projectile's trajectory is ten times larger than the height of the trajectory, then what was the angle of laun

ch with respect to the horizontal? (Assume a flat and horizontal landscape.)
Physics
1 answer:
Lunna [17]3 years ago
3 0

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}

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