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Trigonometry
Question

Suppose a projectile is fired from a cannon with v...

Suppose a projectile is fired from a cannon with velocity \( v _ { 0 } \) and angle of elevation \( \theta \) . The horizontal distance \( R ( \theta ) \) it travels (in feet) is given by the following. 

\( R ( \theta ) = \frac { ( v _ { 0 } ) ^ { 2 } \sin 2 \theta } { 32 } \) 

If \( v _ { 0 } = 80 ft \) /s, what angle \( 0 \) (in radians) should be used to hit a target on the ground \( 118 \) feet in front of the cannon? 

Answer

\(R \theta = 118 \) 

\( V _ { 0 } = 80\) 

\(118 = \frac { 80 ^ { 2 } \sin 2 \theta } { 32 } \)

\(0.59 = \sin 2 \theta \) 

\( \frac { 1 } { 2 } \sin ^ { - 1 } ( 0 ^ { \circ } 59 ) = \theta \) 

\(\theta = 0.316 rad\)

Solution
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