fountains are designed so that the height and distance the water travels is dependent on θ, the angle at which the water is aimed. for any given angle θ, the ratio of maximum height H of the water to the horizontal distance D it travels is given by the formula H/D=1/4/=tanθ. what is the value of θ that will produce a stream of water that goes twice as high as it travels horizontally?
tanA = H/D.
H = 2D.
Substitute 2D for H:
tanA = 2D/D = 2/1 = 2.
A = 63.4 Deg.
To find the value of θ that will produce a stream of water that goes twice as high as it travels horizontally, we need to use the given formula: H/D = tanθ, where H is the maximum height and D is the horizontal distance traveled by the water.
Let's break down the problem step by step:
1. Start with the ratio given in the problem: H/D = 1/2.
This tells us that the maximum height H is twice the horizontal distance D.
So, we have H = 2D.
2. Substitute the value of H in terms of D into the formula H/D = tanθ.
We get (2D)/D = tanθ, which simplifies to 2 = tanθ.
3. Next, we need to find the angle θ that satisfies the equation tanθ = 2.
To do this, we take the inverse tangent (arctan) of both sides of the equation: θ = arctan(2).
4. Finally, use a scientific calculator or an online tool to evaluate the arctan(2).
The value of arctan(2) is approximately 63.4 degrees.
Therefore, the value of θ that will produce a stream of water that goes twice as high as it travels horizontally is approximately 63.4 degrees.