Question

A firefighting crew uses a water cannon that shoots water at 29.0 m/s at a fixed angle of 47.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level.

How far from the building should they position their cannon? There are two possibilities (d1<d2d1<d2); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

Answer 1

Angle projection is  

Height of the target blaze in building is  

Initial velocity is

Horizontal velocity component is  

Initial vertical velocity component is  

In this projectile motion, water will be at the same height at two different points, one while ascending and the other while descending. Let be the distances of these two points from the cannon.

Let x be the horizontal distance covered when water reaches the height h, then

for the horizontal motion, acceleration is 0, therefore, remains the same throughout,

Horizontal distance covered is   ; where t is the time taken, therefore,

for the vertical motion, acceleration due to graviy g is there, therefore

substituting for t, we get

dividing each term by 0.013 and rearranging, we get

from this quadratic equation, we have

These are the two possible distances.