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Veseljchak [2.6K]
4 years ago
14

A shell is fired from the ground with an initial speed of 1.74 ✕ 103 m/s at an initial angle of 58° to the horizontal.

Physics
1 answer:
ValentinkaMS [17]4 years ago
4 0

Answer:

a) horizontal range s=277671.77 m

b) time the shell is in motion 301.143 s

Explanation:

Is a parabolic movement so the velocity have two components:

v = 1.74 x 10^{3}  ( \frac{m}{s} )

\alpha = 58°

v_{y} =v*Sen (\alpha )\\v_{x} =v*Cos (\alpha )

v_{y} =1740*Sen (58 )\\v_{x} =1740*Cos (58)

v_{y} = 1475.603 \frac{m}{s}\\v_{x} = 922.059 \frac{m}{s}\\

t= \frac{2*v_{y} }{g}

t= \frac{2*1475.603 }{9.8}

t= 301.143 s

v= \frac{s}{t} \\s= v_{x}*t

s= 922.059 \frac{m}{s} * 301.143 s

s = 277671.77 m

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