Thursday, December 10, 2009

How far will a rocket travel if it is launched at 1200 km/h at an angle of 50 degrees to the horizontal?

The rocket is launched at an initial speed of 1200 km/h at an angle 50 degrees to the horizontal. As the rocket travels, its movement is opposed by the gravitational force exerted by the Earth and the force of friction due to the air. As there is no information provided about the frictional force, this can be ignored.


The initial velocity of the rocket has vertical and horizontal components equal to 1200*sin 50 km/h and 1200*cos 50 km/h respectively. The deceleration due to gravity can be taken as a constant 9.8 m/s^2 in the downward direction.


As the rocket moves, the vertical component of its velocity decreases. It is equal to 0 at the halfway point. To determine the total horizontal distance traveled by the rocket, first determine the time it travels for. This is equal to 2*T where 0 = 1200*(5/18)*sin 50 - 9.8*T


T = 26.05 s


The horizontal distance traveled is D = 1200*(5/18)*cos 50*2*26.05 = 11163.07 m


The horizontal distance traveled by the rocket is 11163.07 m.

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