Water leaves a fireman’s hose (held near the ground) with an initial velocity v0 = 17.5 m/s at an angle θ = 31° above horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position.

Explanation:

Given that,

Initial velocity U=17.5m/s

Angle of projection is=31°

θ = 31°

We need to find the range of the nozzle water.

The horizontal velocity component of projectile is given as

Vx=UCosθ

Vx=17.5Cos31

Vx=15m/s

Then, the range is given as

R=Vx•t

So we know Vx already, we need to find the time (t), I.e the time of flight.

Using the vertical motion equation of projectile

Initial Vertical velocity component

Voy=USin  θ

Voy=17.5Sin31

Voy=9.013m/s

Then, equation of motion

Vy=Voy-gt.  

Upward motion I.e against gravity

At maximum height , Vy is 0

g =9.81m/s²

O=9.013-9.81t

9.81t=9.103

t=9.103/9.81

t=0.919second

Then apply the range formulas

R=Ux•t

R=15×0.918

R=13.78m

Using the range direct formula

R=U²Sin2θ/2g

R=17.5²Sin(2×31) / (9.81×2)

R=17.5²Sin62/19.62

R=13.78m

So it is correct.

2b2t

explanation:

2b2t