In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 24 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
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To find the new range when the launch speed is doubled, you can use the principles of projectile motion. The range of a projectile depends on factors such as the initial velocity, launch angle, and gravitational acceleration.
First, let's consider the range formula for projectile motion without air resistance:
R = (v0^2 * sin(2θ)) / g
Where:
R is the range
v0 is the initial velocity
θ is the launch angle
g is the acceleration due to gravity
In this case, the initial range is given as 24 m, and we want to find the new range when the initial speed is doubled. Let's denote the new initial speed as 2v0.
Now, we can substitute the new initial velocity into the range formula:
2R = ((2v0)^2 * sin(2θ)) / g
Simplifying the equation further:
2R = (4v0^2 * sin(2θ)) / g
Since the launch angle remains the same, we can cancel out the sin(2θ) term:
2R = (4v0^2 * sin(2θ)) / g = (4v0^2 * 2sinθcosθ) / g
Now, let's divide both sides of the equation by 2:
R = (2v0^2 * sinθ * cosθ) / g
Since sinθ * cosθ = 1/2 * sin(2θ), we can replace sinθ * cosθ with this value:
R = (2v0^2 * 1/2 * sin(2θ)) / g = (v0^2 * sin(2θ)) / g
So, we have:
R = (v0^2 * sin(2θ)) / g
which is the same formula as before without doubling the initial velocity.
Therefore, when the launch speed is doubled and the launch angle remains the same, the new range will remain the same. In other words, the new range will also be 24 m.