a canon is located 200m away from the base of a tower that is 50 m high.?

Question

the canon fires at the speed of 30m/s. find the 2 possible angle of the canon that it will hit the top of the tower.

thanks, i get to the 200 tan θ - 2000/9 sec θ * sec θ = 50 part now but i can't solve for the θ. help please

Answer 1

odu83 is right.
I changed the muzzle velocity to 300 m/s ( a more typical one for a real canon) and the angles turned out to be 14.662° and 89.3744°. These result from using the formula

y = (-g*x²/(2*V²))*sec²Θ + x*tanΘ

and solving for the 2 values of Θ when y = 50m and x = 200m

Answer 2

Let the angle be θ.

At the same instant of time, the projectile travels 200 m horizontally and the 50 m vertically.

The equations of motion are

30 cos θ (t) = 200
30 sin θ (t) - 0.5 * 10 * (t*t) = 50

t = 200/30 cos θ

Substitute this in the second equation and solve.

200 tan θ - 2000/9 sec θ * sec θ = 50

(Use sec^2 θ = 1 + tan ^2 θ)

Answer 3

The cannon ball will never reach an altitude of 50 m even if the cannon fired straight up:
consider the kinetic energy of the ball when fired at 30 m/s
.5*m*30^2
the potential energy when it reaches max height is a complete conversion of the KE, so:
.5*m*30^2=m*g*h
solve for h
.5*900/9.81=45.8 m

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