A projectile is launched at ground level with an initial speed of 51.5 m/s at an angle of 35.0° above the horizontal. It strikes a target above the ground 2.50 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

To solve this problem, we can use the equations of motion for projectile motion. The horizontal and vertical motions are independent of each other, so let's calculate the x and y distances separately.

First, let's find the x distance. The horizontal motion of the projectile is given by the equation:

x = v₀ * cos(θ) * t

Where:
- x is the horizontal distance
- v₀ is the initial speed (51.5 m/s)
- θ is the launch angle (35.0°)
- t is the time (2.50 seconds)

Plugging in the values:

x = 51.5 m/s * cos(35.0°) * 2.50 s

Now, let's calculate the y distance. The vertical motion of the projectile is given by the equation:

y = v₀ * sin(θ) * t + (1/2) * g * t²

Where:
- y is the vertical distance
- v₀ is the initial speed (51.5 m/s)
- θ is the launch angle (35.0°)
- t is the time (2.50 seconds)
- g is the acceleration due to gravity (9.81 m/s²)

Plugging in the values:

y = 51.5 m/s * sin(35.0°) * 2.50 s + (1/2) * 9.81 m/s² * (2.50 s)²

Now that we have the equations, let's calculate the x and y distances.

Calculating the x distance:
x = 51.5 m/s * cos(35.0°) * 2.50 s ≈ 224.35 meters

Calculating the y distance:
y = 51.5 m/s * sin(35.0°) * 2.50 s + (1/2) * 9.81 m/s² * (2.50 s)² ≈ 67.81 meters

Therefore, the x distance from where the projectile was launched to where it lands is approximately 224.35 meters, and the y distance is approximately 67.81 meters.