An arrow is shot at ground level with an initial speed of 50.0 m/s at an angle of 25.0° above horizontal. What is its x-coordinate when the arrow is on the way up and its y coordinate is 5.00 m above the ground?
To find the x-coordinate when the arrow is on the way up and its y-coordinate is 5.00 m above the ground, we can use the equations of motion for projectile motion.
Step 1: Break down the initial velocity into its x and y components.
The initial speed of 50.0 m/s can be broken down into its x and y components using trigonometry. The horizontal component (Vx) is given by Vx = V * cosθ, where V is the initial speed and θ is the angle above the horizontal. Similarly, the vertical component (Vy) is given by Vy = V * sinθ.
Vx = 50.0 m/s * cos(25.0°) = 50.0 m/s * 0.9063 ≈ 45.31 m/s
Vy = 50.0 m/s * sin(25.0°) = 50.0 m/s * 0.423 = 21.15 m/s
Step 2: Use the vertical motion equation to find the time it takes for the arrow to reach a height of 5.00 m above the ground.
The vertical motion equation that relates displacement (y-coordinate), initial velocity (Vy), time (t), and acceleration due to gravity (g) is given by y = Vy * t - 0.5 * g * t^2, where g is approximately -9.8 m/s^2 (negative because it acts downwards).
Rearranging the equation gives:
0 = -0.5 * 9.8 m/s^2 * t^2 + 21.15 m/s * t - 5.00 m
This is a quadratic equation that can be solved for t using the quadratic formula.
Step 3: Solve the quadratic equation to find the time.
Using the quadratic formula t = (-b ± √(b^2 - 4ac)) / 2a, where a = -0.5 * 9.8 m/s^2, b = 21.15 m/s, and c = -5.00 m, we can calculate the time it takes for the arrow to reach a height of 5.00 m above the ground.
The discriminant (b^2 - 4ac) is calculated as:
(21.15 m/s)^2 - 4 * (-0.5 * 9.8 m/s^2) * (-5.00 m) = 447.766 + 98.0 ≈ 545.766
Taking the square root of the discriminant gives:
√(545.766) ≈ 23.36
Using the quadratic formula, we have:
t = (-21.15 m/s ± 23.36) / (-0.5 * 9.8 m/s^2)
For the arrow on the way up, we choose the positive root to get:
t = (-21.15 m/s + 23.36) / (-0.5 * 9.8 m/s^2) ≈ 0.416 s
Step 4: Use the horizontal motion equation to find the x-coordinate.
The horizontal motion equation that relates displacement (x-coordinate), initial velocity (Vx), and time (t) is given by x = Vx * t.
Plugging in the values we have:
x = 45.31 m/s * 0.416 s ≈ 18.85 m
Therefore, the x-coordinate when the arrow is on the way up and its y-coordinate is 5.00 m above the ground is approximately 18.85 meters.