%3Cp%3EA%20cannon%26nbsp%3Bis%26nbsp%3Bfired%26nbsp%3B38%26nbsp%3Bdegrees%26nbsp
ID: 2280358 • Letter: #
Question
%3Cp%3EA%20cannon%26nbsp%3Bis%26nbsp%3Bfired%26nbsp%3B38%26nbsp%3Bdegrees%26nbsp%3Babove%26nbsp%3Bthe%26nbsp%3Bhorizontal%26nbsp%3Bwith%26nbsp%3Ban%26nbsp%3Binitial%26nbsp%3Bspeed%26nbsp%3Bof%26nbsp%3B58m%2Fs%3C%2Fp%3E%3Cp%3Ea%5D%26nbsp%3Bwhat%26nbsp%3Bis%26nbsp%3Bthe%26nbsp%3Bmagnitude%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bhorizontal%26nbsp%3Bcomponent%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bcannon%26nbsp%3Bballs%26nbsp%3Bdisplacement%26nbsp%3Bat%26nbsp%3Bthe%26nbsp%3Bend%26nbsp%3Bof%26nbsp%3B2%26nbsp%3Bsecs%3C%2Fp%3E%3Cp%3Eb%5D%26nbsp%3Bhow%26nbsp%3Blong%26nbsp%3Bdoes%26nbsp%3Bit%26nbsp%3Btake%26nbsp%3Bto%26nbsp%3Breach%26nbsp%3Bits%26nbsp%3Bhighest%26nbsp%3Bpoint%3C%2Fp%3E%3Cp%3Ec%5D%26nbsp%3Bwhat%26nbsp%3Bis%26nbsp%3Bthe%26nbsp%3Bacceleration%26nbsp%3B4%26nbsp%3Bs%26nbsp%3Bafter%26nbsp%3Bit%26nbsp%3Bis%26nbsp%3Bfired%3C%2Fp%3E%3Cp%3Ed%5D%26nbsp%3Bwhat%26nbsp%3Bare%26nbsp%3Bthe%26nbsp%3Bhorizontal%26nbsp%3Band%26nbsp%3Bvertical%26nbsp%3Bcomponants%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bvelocity%26nbsp%3BVx%26nbsp%3Band%26nbsp%3BVfy%26nbsp%3Bat%26nbsp%3Bthe%26nbsp%3Bend%26nbsp%3Bof%26nbsp%3B4s%3C%2Fp%3E%3Cp%3Ee%5Dwhat%26nbsp%3Bis%26nbsp%3Bthe%26nbsp%3Bvelocity%26nbsp%3Bat%26nbsp%3Bthe%26nbsp%3Bend%26nbsp%3Bof%26nbsp%3B4s%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3EPlease%26nbsp%3Bbe%26nbsp%3Bas%26nbsp%3Bdetailed%26nbsp%3Bas%26nbsp%3Bpossible%3C%2Fp%3E%3Cp%3EI%26nbsp%3Bknow%26nbsp%3Bits%26nbsp%3Balot%26nbsp%3Bbut%26nbsp%3BI%26nbsp%3Bwould%26nbsp%3Bappriciate%26nbsp%3Bthe%26nbsp%3Bhelp%3C%2Fp%3EExplanation / Answer
a)
Horizontal distance = Vx * t
= 58*cos(38 deg) * 2
=91.409 m
b)
time taken = Vy / g
t = 58*sin(38 deg) / 9.81
t = 3.64 s
c)
Since there is no acceleration in the horizontal direction, the velocity in the horizontal direction is constant. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to g
a = 9.81 m/s^2
d)
Horizontal Vx = Vxo = 58*cos(38 deg) = 45.705 m/s
Vertical Vy = Vyo - gt = 58*sin(38 deg) - 9.81*4 = -3.5316 m/s
e)
Velocity = 45.705 i - 3.5316 j
Magnitude = sqrt(45.705^2 + 3.5316^2) = 45.84 m/s
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