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An anti-aircraft gun fires at an elevation of 60° at an enemy aircraft at 10000m above the ground At what speed must the cannon be shot to hit the plane at that height?(g=10m/s^2)​

Respuesta :

fichoh

Answer:

516m/s^2

Explanation:

Given the following :

Height of aircraft = 10000m

Acceleration due to gravity (g) = 10m/s^2

Angle of projection (θ) = 60°

Height of aircraft = maximum height

Maximum height of a projectile:

H = (u^2sin^2θ) / 2g

Where H = height

u = initial velocity

10000 = [(usinθ)^2] / 2g

10000 = [(u * sin60°) ^2] / 2*10

10000 = (0.866 * u)^2 / 20

20 * 10000 = 0.749956 * u^2

200000 = 0.749956u^2

u^2 = (200000/0.749956)

u^2 = 266,682.31

u = √266,682.31

u = 516.41292

Initial velocity (u) = 516m/s^2

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