Question

. The height y metres of the water sprayed from a hose , is given as = −
2 + 4 − 1
where x is the horizontal distance travelled by the water. Find the greatest height of the water
sprayed and the horizontal distance from the hose when this occurs?
(PLEASE ANSWER THIS QUICKLY,WHOEVER SENDS AN ANSWER QUICKLY,I WILL MARK THEM AS BRAINLIEST)

Answer 1

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The water strikes a wall 29 meters away at a height 8.18 m

Explanation:

The trajectory of water will be that of a projectile

The angle of projection θ = 60°

Initial velocity u = 20 m/s

Resolving the velocity into two components

The horizontal component of the velocity

u_x=20\cos60^\circ=20\times\frac{1}{2} = 10 \text{ m/s}u

x

=20cos60

∘

=20×

2

1

=10 m/s

And the vertical component of the velocity

u_y=20\sin60^\circ=20\times\frac{\sqrt{3} }{2} = 10\sqrt{3} \text{ m/s}u

y

=20sin60

∘

=20×

2

3

=10

3

m/s

The horizontal distance covered by the stream = 29 m

There is no acceleration in the horizontal direction

Therefore, using distance = speed x time

\begin{lgathered}29=u_x\times t\\\implies 29=10\times t\\\\\implies t=2.9 \text{ sec}\end{lgathered}

29=u

x

×t

⟹29=10×t

⟹t=2.9 sec

Now we need to find the vertical displacement in time 2.9 seconds

using the second equation of motion

h=u_y\times t-\frac{1}{2} gt^2h=u

y

×t−

2

1

gt

2

h=10\sqrt{3} \times 2.9-\frac{1}{2} \times 10\times 2.9^2h=10

3

×2.9−

2

1

×10×2.9

2m

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