. 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?
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Answers
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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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