Asked by jennierson
a projectile with a muzzle velocity of 550 m/s is fired from the a gun on top of a cliff 460 m above the sea level at a certain angle with respect to the horizontal. if the projectile hits the ocean surface. 49.2 seconds after being fired, determine the horizontal range of the projectile?
Answers
Answered by
Henry
Range = Vo^2*2sinA*cosA/g = Vo*cosA*T
(550)^2*2sinA*cosA/9.8=550cosA*49.2
Divide both sides by 550cosA:
550*2sinA/9.8 = 49.2
56.12*2sinA = 49.2
112.24*sin A = 49.2
sinA = 0.43833
A = 26o.
Range = Vo*cos A * T
Range = 550*cos26 * 49.2 = 24321 m.
(550)^2*2sinA*cosA/9.8=550cosA*49.2
Divide both sides by 550cosA:
550*2sinA/9.8 = 49.2
56.12*2sinA = 49.2
112.24*sin A = 49.2
sinA = 0.43833
A = 26o.
Range = Vo*cos A * T
Range = 550*cos26 * 49.2 = 24321 m.
Answered by
atyourworst
this is wrong
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