# Answer to Question #3868 in Other Physics for mia

Question #3868

Water is shooting at the same speed out of the three tubes placed on the ground at different angles of 60°, 45°, and 30° to the horizon. Find the ratio of the greatest heights the streams can reach andthe ratio of distances from the mouths of the tubes the points where the streams hit the ground.

Expert's answer

Let V be the initial velocity of the water.1) The equations of the height are:

H(t) = V*sin(A)*t-(g*t

t

H(t

So H

2) At time tmax water reaches the highest point. So during the time 2*tmax it reaches the ground.

The equation along the horizontal axis is:

S(t) = V*cos(A)*t for t

S

So S

= 3 : 2*sqrt(3) : 3

H(t) = V*sin(A)*t-(g*t

^{2})/2.t

_{max}= V*sin(A)/g.H(t

_{max}) = (V*sin(A))^{2}/g - (V*sin(A))^{2}/(2*g) = (V*sin(A))^{2}/(2*g)So H

_{1max}: H_{2max}: H_{3max}= (V*sin(60°))^{2}/(2*g) : ((V*sin(45°))^{2})/(2*g) : (V*sin(30°))^{2}/(2*g) = sin(60°)^{2}: sin(45°)^{2}: sin(30°)^{2}= 0.75 : 0.5 : 0.25 = 3 : 2 : 12) At time tmax water reaches the highest point. So during the time 2*tmax it reaches the ground.

The equation along the horizontal axis is:

S(t) = V*cos(A)*t for t

_{0}<= t <= 2*t_{max}.S

_{max }= V*cos(A)*2*t_{max}= V*cos(A)*2*V*sin(A)/g = V^{2}*sin(2*A)/gSo S

_{1max}: S_{2max}: S_{3max}= V^{2}*sin(2*60°)/g : V^{2}*sin(2*45°)/g : V^{2}*sin(2*30°)/g = sin(120°) : sin(90°) : sin(60°) == 3 : 2*sqrt(3) : 3

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