Q20E - University Physics with Modern Physics

Question

Firemen use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of 25 m/s as it leaves the end of the hose and then exhibits projectile motion. The firemen adjust the angle of elevation  of the hose until the water takes 3.00 s to reach a building 45 m away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find $\mathit{\alpha }$ . (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

Step-by-Step Solution

Answer

a) The angle of elevation/projection is $53.13°$ .

b) The velocity of water at highest point is 15 m/s and acceleration has magnitude of $9.8 m/s^2$ .

c) The height above the ground is $15.9 m$ .

Step 1: Identification of given data

The given data can be listed below,

• The velocity of water is, $v_0=25 m/s$
• The time taken by water to reach the building is,$t=3.00 s$
• The distance of building from firemen is, x = 45 m

Step 2: Concept/Significance of of projectile motion

A body's motion when thrown at a specific angle from the horizontal with a fixed speed is referred to as projectile motion. The vertical component of a velocity fluctuates whereas the horizontal component always remains constant.

Step 3: (a) Determination of the angle of the elevation.

The horizontal acceleration is zero according to projectile motion and the horizontal speed of the water is constant. the displacement in horizontal direction is given by,

$x=\left(v_0\cos \alpha \right)t$

Here, $\alpha$ is the angle of projection, $v_0$ is the initial speed, and t is the time of duration.

For angle of projection above equation is rearrange as,

$\alpha ={\cos }^{-1}\left(\frac{x}{v_{0}t}\right)$

Substitute all the values in the above,

$$\begin{gathered} \alpha ={\cos }^{-1}\left(\frac{45}{25\times 3}\right) \\ =53.13° \end{gathered}$$

Thus, the angle of elevation/projection is $53.13°$ .

Step 4: (b) Determination of the speed and acceleration of the water at the highest point in its trajectory

The vertical component of velocity will be zero at the highest point in projectile motion so it only have horizontal component of velocity which can be given by,

$v_x=v_0\cos \alpha$

Here, $\alpha$ is the angle of projection, and $v_0$ is the initial speed.

Substitute all the values in the above,

$$\begin{gathered} v_x=\left(25m/s\right)\cos 53.13° \\ =15 m/s \end{gathered}$$

The acceleration of the water only have vertical component acting downward and have magnitude of $9.8 m/s^2$ .

Thus, the velocity of water at highest point is 15 m/s and acceleration has magnitude of $9.8 m/s^2$ .

Step 5: (c) Determination of the height of water strike the building above the ground, and how fast is it moving just before it hits the building.

According to kinematics, the equation for distance of water above the ground is given by,

$y-y_0=v_0\sin \alpha t+\frac{1}{2}a{t}^2$

Here, $y_0$ is the initial position of water, $v_0$ is the initial velocity, t is the flight time, -g is the acceleration in vertical direction.

Substitute all the values in the above,

 

$$\begin{gathered} y=0+\left(25m/s\right)\sin 53.13°\left(3s\right)+\frac{1}{2}\left(-9.8m/s^2\right){\left(3.00s\right)}^2 \\ =15.9 m \end{gathered}$$

Thus, the height above the ground is 15.9 m .