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Ipatiy [6.2K]
3 years ago
13

Topic: Chapter 10: Projectory or trajectile?

Physics
1 answer:
Charra [1.4K]3 years ago
5 0

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

R = \frac{u^2sin(2\theta)}{g}

To find another angle of projection to give the same range;

5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\  the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0

<u>Check: </u>

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

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