I'm having a hard time answering this question. I don't know what equation to use? I've tried using dy= Viyt + 1/2 gt^2. Water is leaving a hose at 6.8 m/s. If the target is 2 m away horizontally, What angle should the water have initially? thank you :D
Range = (Vo^2/g)*sin(2A) = 2.0 m
Vo = 6.8 m/s is the initial velocity.
Solve for A.
sin(2A) = 0.4239
2A = 25.08 or 154.92 degrees
A = 12.54 or 77.46 degrees
Thank YOU :D
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I never thought I'd see that name again.
To solve this problem, we can use the equations of motion for projectile motion. The equation you mentioned, dy = Viyt + 1/2gt^2, relates the vertical distance covered (dy), initial vertical velocity (Viy), time (t), and acceleration due to gravity (g).
However, since we are looking for the initial angle at which the water should be launched, we need to use the horizontal and vertical components of the velocity separately.
Let's break this problem into parts:
1. Find the time of flight: The time for the water to reach the target can be found using the horizontal velocity. The horizontal distance (dx) can be related to the initial horizontal velocity (Vix) and time (t) using the equation dx = Vix * t.
Given that the distance (dx) is 2 m and the horizontal velocity (Vix) is 6.8 m/s, we can rearrange the equation to solve for time:
t = dx / Vix
2. Find the vertical component of the velocity (Viy): Using the equation dy = Viyt + 1/2gt^2, we know that when the water reaches the target (dy = 0), time (t) is the time of flight, and acceleration due to gravity (g) is -9.8 m/s^2 (considering the downward direction as positive).
Rearranging the equation, we get:
-1/2gt^2 = Viyt
Since dy = 0 at the target, we can write it as:
0 = Viyt + 1/2gt^2
Substituting values, we have:
0 = Viy * (dx / Vix) + 1/2 * (-9.8) * (dx / Vix)^2
3. Find the initial angle: The initial angle (θ) can be found using the relationship between the vertical and horizontal components of the velocity. Remember, tan(θ) = Viy / Vix.
Rearranging the equation and substituting Viy, we get:
tan(θ) = (1/2 * (-9.8) * (dx / Vix)^2) / (dx / Vix)
Simplifying, we have:
tan(θ) = -4.9 * (dx / Vix)
Finally, solve for θ by taking the inverse tangent (tan^-1) of both sides:
θ = tan^-1(-4.9 * (dx / Vix))
Plugging in the values dx = 2 m and Vix = 6.8 m/s, you can calculate the value of θ.
By following these steps and equations, you can determine the angle at which the water should be initially launched to hit the target.