Question

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation α of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.

a) Find the angle of elevation αb) Find the speed and acceleration of the water at the highestpoint in its trajectoryc) How high above the ground does the water strike thebuilding, and how fast is it moving just before it hits thebuilding?

Answer 1

Answer and Solution:

As per the question:

Velocity of water,
`v_(o) = 25.0\ m/s`

Time taken by the water to reach the building, t = 3.00 s

Distance of the building, d = 45.0 m

Now,

(a) For the angle of elevation of the hose,
`\alpha`:

`d = v_(o)tcos(\alpha)`

`45 = 25* 3cos(\alpha)`

`45 = 75cos(\alpha)`

`\alpha = cos^(- 1)(0.6) = 53.13^(\circ)`

(b) At the highest point in the trajectory:

Speed =
`v_(o)cos\alpha = 25cos53.13^(circ) = 15\ m/s`

Acceleration, a is due to gravity acting vertically downwards;

a = - g = 9.8
`m/s^(2)`

(c) The height at which the water strikes the building is given by:

H =
`v_(o)sin\alpha t - (1)/(2)gt^(2)`

H =
`25.0* 3sin(53.13^(\circ)) - (1)/(2)* 9.8* 3^(2)`

H = 15.89 m

Now, the vertical component of the velocity is v' at t = 3.00 s is given by;

v' =
`v_(o)sin\alpha - gt`

v' =
`25sin(53.13^(\circ)) - 9.8* 3 = - 9.4 m/s`

Now, the resultant velocity from the horizontal and vertical component is given by:

v =
`\sqrt{v_(o)^(2) + v'^(2)} = \sqrt{25^(2) + (- 9.4)^(2)} = 26.71\ m/s`