A dart gun is fired while being held horizontally at a height of 1.52 m above ground level, and at rest relative to the ground. The dart from the gun travels a horizontal distance of 7.81 m. A child holds the same gun in a horizontal position while sliding down a
35.7◦ incline at a constant speed of 0.807 m/s.
A.) What horizontal distance x will the dart travel if the child fires the gun forward when it is 0.952 m above the ground? The acceleration due to gravity is 9.8 m/s^2. Answer in units of m
To solve this problem, we can use the kinematic equations for projectile motion. In this case, the dart is being fired horizontally, so the initial vertical velocity and the final vertical velocity will both be equal to zero.
Given:
Initial vertical height (y₀) = 1.52 m
Horizontal distance traveled (x) = 7.81 m
Acceleration due to gravity (g) = 9.8 m/s²
Final vertical height (y) = 0.952 m
First, we need to find the time it takes for the dart to reach the final height. We can use the equation:
y = y₀ + v₀y * t + (1/2) * g * t²
Since y₀ = y and v₀y = 0, the equation simplifies to:
0.952 = 1.52 + (1/2) * 9.8 * t²
Rearranging the equation, we get:
0 = 4.9 * t² - 9.6 * t + 0.568
Solving this quadratic equation, we find two possible solutions for t: t₁ ≈ 0.444 s and t₂ ≈ 1.188 s.
Since we're interested in the time it takes for the dart to be at a height of 0.952 m, we choose t = t₁ ≈ 0.444 s.
Now, we can find the horizontal distance traveled by the dart. We use the equation:
x = v₀x * t
We know that the vertical velocity (v₀y) is zero, so the horizontal velocity (v₀x) remains constant. Let's calculate it:
v₀x = x / t
= 7.81 m / 0.444 s
≈ 17.58 m/s
Finally, we can use this velocity to find the horizontal distance (x) when the child fires the gun:
x = v₀x * t
= 17.58 m/s * 0.807 s
≈ 14.20 m
Therefore, the horizontal distance (x) that the dart will travel when the child fires the gun at a height of 0.952 m is approximately 14.20 m.
To find the horizontal distance that the dart will travel when the child fires the gun, we can use the projectile motion equations.
First, let's find the initial velocity of the dart. Since the child is sliding down the incline at a constant speed, the horizontal component of the child's velocity will be the same as the initial velocity of the dart. We can find this using trigonometry.
The horizontal component of the child's velocity (Vx) can be found using the formula:
Vx = velocity * cos(angle)
Given:
velocity = 0.807 m/s
angle = 35.7 degrees
Substituting the values into the formula, we get:
Vx = 0.807 m/s * cos(35.7 degrees)
Vx = 0.807 m/s * 0.819
Vx = 0.661 m/s
Now, let's find the time it takes for the dart to reach the ground. We can use the height (h) and the acceleration due to gravity (g) to calculate this. The formula to find the time is:
h = (1/2) * g * t^2
Rearranging the formula to solve for t, we get:
t = sqrt((2 * h) / g)
Given:
h = 0.952 m
g = 9.8 m/s^2
Substituting the values into the formula, we get:
t = sqrt((2 * 0.952 m) / 9.8 m/s^2)
t = sqrt(0.194 m / s^2)
t ≈ 0.441 s
Now that we know the time it takes for the dart to reach the ground, we can find the horizontal distance it travels (x) using the formula:
x = Vx * t
Substituting the values we calculated earlier, we get:
x = 0.661 m/s * 0.441 s
x ≈ 0.292 m
Therefore, the horizontal distance (x) that the dart will travel when the child fires the gun at a height of 0.952 m above the ground is approximately 0.292 m.