Question

Topic: Chapter 10: Projectory or trajectile?

Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?

90°
75°
45
50°
30°

Answer 1

The other angle is 75⁰

Step-by-step 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 = (u^2sin(2\theta))/(g)`

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

`5.1 = (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`

Check:

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

sin(2θ) = sin(2 x 15) = sin(30) = 0.5