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4 years ago

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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$\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)$

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$\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)$

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