A projectile is launched at an angle of 45 degrees to the horizontal with an initial speed of 200 m/s. Consider the variation in air density with altitude, and determine the maximum height and range of the projectile.
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To determine the maximum height and range of the projectile, we need to consider the effect of air density on its motion.
The maximum height can be found by considering the vertical motion of the projectile. The initial velocity in the vertical direction is given by:
Vy = V * sin(θ)
= 200 m/s * sin(45°)
= 141.42 m/s
The time taken to reach the maximum height can be found using the equation:
Δy = Vy * t - 0.5 * g * t^2
where Δy is the displacement in the vertical direction, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
At the maximum height, the displacement in the vertical direction is zero:
0 = 141.42 m/s * t - 0.5 * 9.8 m/s^2 * t^2
Simplifying the equation and solving for t, we get:
4.9t^2 - 141.42t = 0
t(4.9t - 141.42) = 0
t = 0 (ignoring)
4.9t - 141.42 = 0
t ≈ 28.88 s
The maximum height (H) can be found using the equation:
H = Vy * t - 0.5 * g * t^2
H = 141.42 m/s * 28.88 s - 0.5 * 9.8 m/s^2 * (28.88 s)^2
H ≈ 4093.47 m
Therefore, the maximum height of the projectile is approximately 4093.47 meters.
To determine the range, we consider the horizontal motion of the projectile. The initial velocity in the horizontal direction is given by:
Vx = V * cos(θ)
= 200 m/s * cos(45°)
= 141.42 m/s
The time of flight (T) can be found using the equation:
T = 2 * t
= 2 * 28.88 s
= 57.76 s
The range (R) can be found using the equation:
R = Vx * T
= 141.42 m/s * 57.76 s
= 8166.46 m
Therefore, the range of the projectile is approximately 8166.46 meters.
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