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Home Q & A Board Science-Related Homework Help Physics
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judd judd
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11 years ago
A projectile is fired with an initial speed of 46.6  at an angle of 45.2 above the horizontal on a long flat firing range.
Determine the speed of the projectile 1.50  after firing.
Determine the direction of the motion of the projectile 1.50  after firing.
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nutrition_mannutrition_man
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11 years ago
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#2
11 years ago
Vy = Uy - gt = 46.6*sin(45.2) - 9.8*1.5 = 18.36 mps.
Vx = Ux = U cos(45.2)  = 46.6*cos(45.2) =32.83 mps

V = sqrt(Vx^2 + Vy^2) = sqrt(32.83^2 + 18.36^2) = 37.6 mps at theta = ATAN(Vy/Vx) = ATAN(18.36/32.83) = 29.2  deg above the X axis.  ANS.

NOTE:  The angle is above, not below, the X axis as Vy > 0, pointing up; the projectile is still climbing.  It's be a few more seconds before the projectile starts back down.
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