A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53 degrees above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1

8.8 m or 52.5 m from the building

Explanation:

The water cannon shoots at 25m/s at a fixed angle 53 degrees above the horizontal; the velocity of the water will have vertical components and horizontal component. The vertical component his responsible for the vertical motion of the water.

vertical components of U = U sin 53oc = 25 sin 53oc = 0.7986 * 25 = 19.96 m/s where U is the initial velocity of the water

using equation of motion and S = height = 10 m

S = ut + 1/2 at² a = g = - 9.81 m/s^2 since it act downward

10 = 19.97t + 1/2 * - 9.81 t² = 19.95t - 4.9t²

bring 10 in to form a quadratic equation and multiply equation by (-1)

4.9t² - 19.97t + 10 = 0

solve using quadratic formula

-b±√ (b² - 4ac) / 2a where a = 4.9 b = 19.97 and c = - 10

-19.97±√ (398.8 - 196) / (2*4.9)

-19.97±√ (202.8) / 9.8

(-19.97±14.24) : 9.8

-34.21/9.8 or - 5.73/9.8

t = - 3.49 or t = - 0.585s

solving for the horizontal motion

horizontal distance = horizontal component of the velocity * t since it will travel both at the same time

horizontal distance = 25 cos 53oc * 0.585 = 8.8m from the building or

horizontal distance = 25 cos 53oc * 3.49 = 52.5m from the building