A projectile is fired at v0= 381.0 m/s at an angle of theta = 69.3° with respect

Question

A projectile is fired at v0= 381.0 m/s at an angle of theta = 69.3° with respect to the horizontal. Assume that air friction will shorten the range by 29.1%. How far will the projectile travel in a horizontal direction, R? Sapling Learning Map macnillan learning A projectile is fired at Vo = 3810 m/s at an angle of -69.30 with respect to the horizontal. Assume that air friction will shorten the range by 29.1%. How far will the projectile travel in the horizontal direction, R? Number km Vo Exit Previous Give Up & View Solution Q Check Answer e Next-I Hint

Explanation / Answer

  Horizontal V component = (cos 69.3) x 381 = 134.32 m/sec.
Initial vertical V component = (sin 69.3) x 381 = 355.46 m/sec.
Time to max. height, no friction = (v/g) = 355.46/9.8, = 36.27 secs.
(36.27 x 2) x 134.32 = 9743.57 metres, no friction.
( 9743.57/100) x (100 - 29.1) =6908.19 metres.

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