A projectile is launched with speed 20 m/s at 45 degrees. What is its range neglecting air resistance?

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Multiple Choice

A projectile is launched with speed 20 m/s at 45 degrees. What is its range neglecting air resistance?

Explanation:
When air resistance is neglected, the horizontal range of a projectile depends on its speed and launch angle through the range formula R = v^2 sin(2θ) / g. This comes from breaking the motion into horizontal and vertical parts: horizontal speed is v cosθ, vertical speed is v sinθ, and the flight time is 2v sinθ / g, giving R = v cosθ × (2v sinθ / g) = v^2 sin(2θ) / g. Here, the speed is 20 m/s and the angle is 45°. Since sin(2×45°) = sin(90°) = 1, the range is R = 20^2 / g = 400 / 9.8 ≈ 40.8 m. So the projectile lands about 40.8 meters away.

When air resistance is neglected, the horizontal range of a projectile depends on its speed and launch angle through the range formula R = v^2 sin(2θ) / g. This comes from breaking the motion into horizontal and vertical parts: horizontal speed is v cosθ, vertical speed is v sinθ, and the flight time is 2v sinθ / g, giving R = v cosθ × (2v sinθ / g) = v^2 sin(2θ) / g.

Here, the speed is 20 m/s and the angle is 45°. Since sin(2×45°) = sin(90°) = 1, the range is R = 20^2 / g = 400 / 9.8 ≈ 40.8 m. So the projectile lands about 40.8 meters away.

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