Question

A field gun is pointed at an angle of 35° above the horizontal and fires a shell with a muzzle velocity of 650m/s. Ignoring air resistance,to what peak height will the shell rise and what will be it's range?
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Answer 1

Given:

A field gun is pointed at an angle of 35° above the horizontal and fires a shell with a muzzle velocity of 650m/s.

To find:

• Peak height

• Range of shell

Calculation:

Let peak height be h ;

$\therefore \: h = \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{2g}$

$= > \: h = \dfrac{ {(650)}^{2} \times { \sin}^{2} ( {35}^{ \circ} ) }{2 \times 10}$

$= > \: h = \dfrac{ {(650)}^{2} \times 0.32 }{2 \times 10}$

$= > \: h = \dfrac{ {(650)}^{2} \times 0.16 }{ 10}$

$= > \: h \approx \: 6760 \: m$

$\boxed{= > \: h \approx \: 6.76 \: km}$

Let range be R ;

$\therefore \: R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$= > \: R = \dfrac{ {(650)}^{2} \sin( {70}^{ \circ} ) }{10}$

$= > \: R \approx \: 39702.01 \: m$

$\boxed{= > \: R \approx \: 39.7 \: km}$

Hope It Helps.