Physics projectile problem

[lmf22]

physics projectile problem...

A fire hose held near the gound shotts water at a speed of 6.8m/s. At what angle(s) should the nozzle point in order that the water would land 3.0m away?
It asks for 2 different angles.

Any help would be great. Thank you.

 

[ShawnD]

Make an equation showing the time spent in the air using the distance formula. This is overall vertical distance (0).

d = Vi*t + (1/2)*a*t^2

d is 0, Vi should be 6.8*sin(theta), a is just -9.80. Try to isolate the time. From the looks of it, the time will equal something * sin(theta).

From there, do horizontal distance

d = Vi*t

d is 3, Vi is 6.8*cos(theta), and t is the equation created above. You should end with sin(theta)*cos(theta) = something. From there you can probably just graph it.

 

[M98Ranger]

To solve this physics projectile problem, we can use the equations of motion for a projectile:

1. Horizontal displacement (x) = initial velocity (v) * time (t) * cos(theta)
2. Vertical displacement (y) = initial velocity (v) * time (t) * sin(theta) - (1/2) * acceleration due to gravity (g) * time (t)^2
3. Final velocity in the horizontal direction (vx) = initial velocity (v) * cos(theta)
4. Final velocity in the vertical direction (vy) = initial velocity (v) * sin(theta) - g * time (t)

Given:
Initial velocity (v) = 6.8m/s
Horizontal displacement (x) = 3.0m
Vertical displacement (y) = 0m (since the water is landing on the ground)
Acceleration due to gravity (g) = 9.8m/s^2

Substituting these values into the equations, we can solve for the time (t) and the angle (theta).

1. x = 6.8 * t * cos(theta)
2. y = 6.8 * t * sin(theta) - 4.9 * t^2
3. 3.0 = 6.8 * t * cos(theta)
4. 0 = 6.8 * t * sin(theta) - 4.9 * t^2

Solving for t in equation 3:
t = 3.0 / (6.8 * cos(theta))

Substituting this value of t into equation 4:
0 = 6.8 * (3.0 / (6.8 * cos(theta))) * sin(theta) - 4.9 * (3.0 / (6.8 * cos(theta)))^2

Simplifying:
0 = 3.0 * tan(theta) - 4.9 * 0.132 * sec^2(theta)

Using a calculator or graphing the equation, we can find that there are two possible angles:
theta = 32.6 degrees and theta = 57.4 degrees

Therefore, the nozzle should point at an angle of 32.6 degrees and 57.4 degrees in order for the water to land 3.0m away.