A cannon with a muzzle speed of 990 m/s is used to start an avalache on a mounta

Question

A cannon with a muzzle speed of 990 m/s is used to start an avalache on a mountain slope. The target is 1950 m from the cannon horizontally and 804 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.) 21.31 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. Need Help? Read It

Explanation / Answer

let the angle be theta.

let initial horizontal speed=990*cos(theta) m/s

and initial vertical speed=990*sin(theta) m/s

let time taken to reach 804 m height is t.

then horizontal distance is also covered in time t

==> 990*cos(theta)*t=1950

==>cos(theta)*t=1.9696

==> t=1.9696*sec(theta)....(1)

and vertical height covered=804 m

==>990*sin(theta)*t-0.5*9.8*t^2=804

==>990*sin(theta)*1.9696*sec(theta)-0.5*9.8*1.9696^2*sec^2(theta)=804

==>1950*tan(theta)-19*sec^2(theta)=804

==>1950*tan(theta)-19*(1+tan^2(theta))=804

==>1950*tan(theta)-19*tan^2(theta)-823=0

solving for tan(theta), we get

tan(theta)=0.423

==>theta=22.928 degrees

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