A projectile has a range 80 m and reaches a maximum height of 20 m. The angle of elevation at which the projectile is fired is
Question
A projectile has a range 80 m and reaches a maximum height of 20 m. The angle of elevation at which the projectile is fired is
Solution
The range (R) and the maximum height (H) of a projectile are given by the following formulas:
R = (v^2 * sin(2*theta)) / g
H = (v^2 * sin^2(theta)) / (2*g)
where:
- v is the initial velocity,
- theta is the angle of elevation,
- g is the acceleration due to gravity.
Given that R = 80 m and H = 20 m, we can use these two equations to solve for theta.
First, divide the first equation by the second to get:
R / H = 2 * cot(theta)
Substituting the given values:
80 / 20 = 2 * cot(theta)
4 = cot(theta)
Therefore, theta = cot^-1(4)
Using a calculator, we find that theta is approximately 14.04 degrees.
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