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.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.