Question

a pipe lying horizontally on ground pumps water towards a wall 8 meters away from the end of the pipe with a velocity of 20 m s -1 if the water stream makes an angle of 37 degree with the horizontal then height at which water will hit the wall will be?​

4.75m
3.25m
6m
2.8m
​

Answer 1

Answer:

a. 4.75m

Explanation:

we know that ux= ucosQt

thus, 20×cos37×t= 8

20×4/5×t=8

t=1/2=0.5

now,

using uy=usinQt-1/2gt²

uy= 20× sin37° ×0.5 - 1/2×10×0.25

uy= 6- 1.25

uy= 4.75m

hence, the hight will be 4.75m

hope it helps:)

Answer 2

Given:

A pipe lying horizontally on ground pumps water towards a wall 8 meters away from the end of the pipe with a velocity of 20 m/s.

To find:

If the water stream makes an angle of 37° with the horizontal then height at which water will hit the wall will be?

Calculation:

The general equation of trajectory of a projectile is as follows :

$\boxed{ \bold{ \therefore \: y = x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2}( \theta) } }}$

• y is the instantaneous height of the projectile from the ground.

• x is the instantaneous horizontal distance from the point of projection.

• u is the initial velocity.

• $\theta$ is the angle of projection.

Putting the available values:

$\sf \therefore \: y = x \tan( \theta) - \dfrac{ g{x}^{2} }{2 {u}^{2} { \cos}^{2}( \theta) }$

$\sf \implies \: y = 8\tan( {37}^{ \circ} ) - \dfrac{g {(8)}^{2} }{2 \times {(20)}^{2} { \cos}^{2}( {37}^{ \circ} ) }$

$\sf \implies \: y =( 8 \times \dfrac{3}{4} ) - \dfrac{g {(8)}^{2} }{2 \times {(20)}^{2} {( \frac{4}{5} )}^{2} }$

$\sf \implies \: y = 6 - \dfrac{g\times64 \times 25}{2 \times 400 \times 16}$

$\sf \implies \: y = 6 - \dfrac{g\times64}{2 \times 16 \times 16}$

$\sf \implies \: y = 6 - \dfrac{g\times4}{2 \times 16}$

$\sf \implies \: y = 6 - \dfrac{10}{8}$

$\sf \implies \: y = 6 - 1.25$

$\sf \implies \: y = 4.75 \: m$

So, the water will hit the wall at a height of 4.75 metres.