Question

a water pipe lying on ground has a hole in it. from this hole water stream shoots upward at an angle 30 to horizontal. the speed of water stream is 10 m/s. at what height stream hit the wall which is 6 m away from hole?​

Answer 1

solution is above the answer is 6.78m

Answer 2

The stream will hit the wall at 1.064 m.

Given,

Angle of projection = θ = 30°

Speed of projection = u = 10 m/s

Distance of wall from hole = 6 m

To Find,

Height at which the stream of water hits the wall

Solution,

The given situation represents a projectile motion,

Firstly, we need to check whether the wall is in the range of the projectile stream of water

Let the range be 'R'

R = (u²Sin2θ)/g

where g is the acceleration due to gravity

u is projection speed

and θ is the angle of projection

R = $\frac{10*10*Sin(2*30)}{10}$

R = $\frac{10*10*Sin(60)}{10}$

R = 10 * Sin 60°

R = 10 * $\frac{\sqrt{3} }{2}$

R = 5√3 ≈ 8.66 m

Thus, the wall lies in the range of the projectile.

Now, to find the height at which the water stream hits the wall, we can use the equation of trajectory which is given as -

y = xTanθ - $\frac{gx^2}{2u^2cos^2theta}$

where y is the vertical distance and x is the horizontal distance

x = 6 m (since the wall is at 6 m from the hole)

y = 6 * Tan 30° - $\frac{10 * 6 *6}{2*10*10*cos^230}$

y = 6 * $\frac{1}{\sqrt{3} }$ - $\frac{360}{200*(\frac{\sqrt{3} }{2} )^2}$

y = $\frac{6}{\sqrt{3} }$ - $\frac{360}{200*\frac{3} {4}}$

y = $\frac{6}{\sqrt{3} }$ - $\frac{360}{150}$

y = 3.464 - 2.400

y = 1.064 m

Thus, the stream will hit the wall at 1.064 m.

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