A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizontal distance that the dart can travel? Assume that the dart gun is fired at ground level.
Answers:
A) 6.14 m
B) 3.88 m
C) More information is needed.
D) 12.28 m
Thank you to whoever helps with this
max range = v^2 / g
To find the maximum horizontal distance that the dart can travel, we need to understand the projectile motion and the concept of range. The range is the horizontal distance covered by a projectile before hitting the ground. In this case, the dart is fired at ground level, so the range will give us the maximum horizontal distance.
The formula for the range (R) of a projectile is given as:
R = (v^2 * sin(2θ)) / g
Where:
- v is the initial velocity of the projectile (7.76 m/s in this case)
- θ is the angle of projection
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
However, in this question, it is provided that the initial velocity remains the same regardless of the angle of projection. This means that the angle of projection does not affect the magnitude of the velocity but only the direction. So, to maximize the range, we need to find the optimum angle that maximizes the value sin(2θ).
The maximum value of sine function is 1, which occurs at 90 degrees. Therefore, the optimum angle for maximum range is 45 degrees.
Now, substituting the values in the range formula:
R = (7.76^2 * sin(2 * 45)) / 9.8
Simplifying:
R = (60.0976 * 1) / 9.8
R ≈ 6.14 m
The maximum horizontal distance that the dart can travel is approximately 6.14 meters.
Therefore, the correct answer is (A) 6.14 m.
To find the maximum horizontal distance that the dart can travel, we can use the formula for the range of projectile motion. The range is the horizontal distance traveled by the projectile before it hits the ground.
The formula for the range (R) is given by:
R = (V^2 * sin(2θ)) / g
where V is the initial velocity of the dart, θ is the launch angle, and g is the acceleration due to gravity.
In this case, the initial velocity (V) is given as 7.76 m/s, and the dart is fired at ground level, so we assume the launch angle (θ) to be 0 degrees.
Let's calculate the range using these values:
R = (7.76^2 * sin(2 * 0)) / g
Since sin(0) is 0, we can simplify the equation further:
R = (7.76^2 * sin(0)) / g
R = 0 / g
R = 0
Based on the calculations, the maximum horizontal distance that the dart can travel is 0 meters. Therefore, the correct answer from the given options is:
C) More information is needed.