Question

WILL BE MARKED AS BRAINLIEST FOR CORRECT ANSWER WITH EXPLANATION!

A boy while watering a garden keeps outlet of the hose h = 0.8 m above the ground.
Water is continuously flowing out of the hose with a constant velocity u=6m/s at an angle Ф=30° above the horizontal.Cross sectional area of the outlet A=1.5$m^{2}$
,density of the water is 1000 kg/m^3 and the acceleration of the free fall is g=10m/s.
Find the mass of the water in the water stream.

Answer 1

so we basically need to find the length of the path taken by the stream of water or the time taken by the first ring of water from the hose to hit the ground t.e the projectile time.

let x = initial speed along  y-axis = 6 x sin(30) = 3

time to reach the ground = t1 + t2

t1 -> time to reach the same height point in return journey
 
t1 = (-3-3)/(-10) = 0.6 sec    (v = u+at) (took downwards as negative)

s = ut + (1/2) a (t^2)            (t^2 -> t square)

s = 0.8m

so the equation becomes 5(t^2) - 6t + 0.8 = 0

solve it and get the t which is t2

so then u get the total time

after getting the total time we can find the required volume of water as now we know the path length which is
(total time) x (velocity of the water hose) = dist


volume of water = dist  * 1.5

now
volume of water * density of water = required mass of the water

Answer 2

y component of velocity=u*sin30=6*1/2=3m/s

The outlet of hose is 0.8m above ground,so from that point,water falls as a projectile with y component of velocity undergoing free fall.

s=ut+1/2at^2
0.8=3t+1/2*10*t^2
 
5t^2+3t-0.8=0

(t-0.2)(5t+4)=0
t=0.2s,-0.8s
water retains in pipe for 0.2s.
cross section area*speed of water flow*retention time=amt of water
1.5m^2*6m/s*0.2s=1.8m^3
density*volume=mass of water
1000kg/m^3*1.8m^3=1800kg
answer is 1800 kg..