# physics

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Short projectile motion question

A German rocket from the second world war had a range of 3000meters reaching a maximum height of 1000meters.
Determine the rockets maximum velocity

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Vertical height h = V^2(sin^µ)/2g.
Horizontal distance = d = V^2(sin2µ)/g.
h = height, meters V = initial launch velocity, m/s, g = acceleration due to gravity, m/s^2s and µ = the elevation angle of the rocket measured from the horizontal to the launch directiion.

Thus, h = V^2(sin^2µ)/2g = 1000 and
......d = V^2(sin2µ)/g = 3000.

Solve for V^2, equate the results and solve for µ and then V.

Ignoring atmospheric drag and the influence of gravity.

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