Prove by making calculations that a cannon will shoot three times as high when the angle of projection is 60 as when it is 30°.
To prove this, we can use the equation for the maximum height achieved by a projectile:
h = (v^2 * sin^2(theta)) / (2 * g),
where:
h is the maximum height,
v is the initial velocity of the projectile,
theta is the angle of projection,
and g is the acceleration due to gravity.
Let's assume the initial velocity v and the acceleration due to gravity g are constant for both angles of projection (30° and 60°). We can compare the heights achieved at 30° and 60° by dividing one equation by the other:
h(60) / h(30) = ((v^2 * sin^2(60°)) / (2 * g)) / ((v^2 * sin^2(30°)) / (2 * g)),
h(60) / h(30) = (sin^2(60°)) / (sin^2(30°)).
Let's calculate this ratio:
h(60) / h(30) = (0.866)^2 / (0.5)^2,
h(60) / h(30) = 0.75 / 0.25,
h(60) / h(30) = 3.
Therefore, a cannon will shoot three times as high when the angle of projection is 60° compared to when it is 30°.