A bullet is fired at an angle of 30° above the horizontal with a velocity of 500m/s.
a)find the range.
b)Solve the time of itsflight.
c)At what other qngle of elevation coyld this bullet be fired to give the same range as in(a).
What is the maximum height attained by the bullet?
(a,c) see the article on Trajectory in wikipedia
(b) the height h is
h(t) = (500 sin30°) t - 4.9t^2
so solve for t when h=0
To find the range (horizontal distance traveled by the bullet), time of flight, and maximum height attained by the bullet, you can use the following equations of motion:
1. Range (R): R = (v^2 * sin(2θ)) / g
2. Time of flight (T): T = (2 * v * sin(θ)) / g
3. Maximum height (H): H = (v^2 * sin^2(θ)) / (2 * g)
where v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity (approximately 9.8 m/s²).
a) Range (R):
Plug in the given values: v = 500 m/s and θ = 30°, and g = 9.8 m/s² into the range equation:
R = (v^2 * sin(2θ)) / g
= (500^2 * sin(2 * 30°)) / 9.8
= (250000 * sin(60°)) / 9.8
= (250000 * √3/2) / 9.8
≈ 3387.76 m
b) Time of flight (T):
Plug in the given values into the time of flight equation:
T = (2 * v * sin(θ)) / g
= (2 * 500 * sin(30°)) / 9.8
= (1000 * 0.5) / 9.8
≈ 51.02 s
c) To find the angle of elevation that gives the same range as in part (a), you need to solve the range equation for θ. Rearranging the equation, we have:
sin(2θ) = (R * g) / (v^2)
2θ = arcsin((R * g) / (v^2))
θ = (1/2) * arcsin((R * g) / (v^2))
Substituting the values from part (a), we get:
θ = (1/2) * arcsin((3387.76 * 9.8) / (500^2))
≈ 24.51°
Therefore, the bullet could be fired at an angle of elevation of approximately 24.51° to give the same range.
d) Maximum height (H):
Plug in the given values into the maximum height equation:
H = (v^2 * sin^2(θ)) / (2 * g)
= (500^2 * sin^2(30°)) / (2 * 9.8)
= (250000 * 0.25) / 19.6
≈ 318.88 m
So, the maximum height attained by the bullet is approximately 318.88 meters.