Question

a field is on the slope of a hill of inclination45° with horizontal .for irrigation purpose a vertical pole us placed somewhere on the field ,which throws a water jet with velocity v0 horizontally in the direction shown.the plane of motion includes the line of maximum slope on the field .if the pole is 15m high and the water strikes the field perpendicularly.find value of v0: options are ..(a)5m/s (b) 7.5m/s (C) 10m/s (D) 15m/s
plz guys solve this question its very urgent .no need to give me useless answers​

Answer 1

Explanation:

On striking the plane at B, the velocity along the x-axis should be zero. So using v

x =u x +a x t, we get 0=v 0

cosα−gsinβT, where T is the time of flight

T= gsinβ v 0 cosα

....(i)Also S y

=0 from A to B. So using S y

=u y t+ 21

a y t 2

, we get

0=(v 0 sinα)T− 21

(gcosβ)T 2 ⇒T= gcosβ2v 0sinα

From (i) and (ii),

gcosβv 0 cosα = gcosβ 2v 0

sinα ⇒tanα= 21

cotβ⇒cosα= 4+cot 2 β

2 = 1+3sin 2β2sinβ

Put the value of cosa i

(i) to get T= g 1+3sin 2 β2v 0

Since the body collides elastically, it will rebound perpendicular to the inclined plane with the same speed v. In consequence, it retraces its path elapsing equal time T for backward journey to the point of projection. Hence, the required time of motion is

T ′ =2T= g 1+3sin 2 β 4v 0

hope you gotcha

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