Answer:
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
