A projectile is launched at ground level with an initial speed of 56.0 m/s at an angle of 35.0° above the horizontal. It strikes a target above the ground 2.90 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Vo = 56m/s[35o].

Xo = 56*Cos35 = 45.9 m/s.
Yo = 56*sin35 = 32.1 m/s.

Y = Yo*t + 0.5g*t^2=32.1*2.9 -4.9*2.9^2
= 52 m.

X = Xo*T = 45.9m/s * 2.9s = 133 m.

To find the x and y distances from where the projectile was launched to where it lands, we can use the equations of motion for projectile motion.

First, we can analyze the vertical motion of the projectile. We know that the initial vertical velocity is given by:
vy_initial = v_initial * sin(theta)
where v_initial is the initial speed of the projectile and theta is the launch angle.

So, substituting the given values, we have:
vy_initial = 56.0 m/s * sin(35.0°)
= 56.0 m/s * 0.5736
= 32.118 m/s

Next, we can use the equation for vertical displacement:
Δy = vy_initial * t + (1/2) * g * t^2
where Δy is the vertical displacement, vy_initial is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.

We are given that the time is 2.90 seconds and the acceleration due to gravity is approximately 9.8 m/s^2. Substituting these values, we have:
Δy = 32.118 m/s * 2.90 s + (1/2) * 9.8 m/s^2 * (2.90 s)^2
= 93.0798 m + 40.0398 m
= 133.1196 m

Therefore, the vertical distance from where the projectile was launched to where it lands is 133.1196 meters.

Moving on to the horizontal motion, we can use the equation for horizontal displacement:
Δx = vx_initial * t
where Δx is the horizontal displacement and vx_initial is the initial horizontal velocity.

The initial horizontal velocity can be calculated as:
vx_initial = v_initial * cos(theta)
where v_initial is the initial speed of the projectile and theta is the launch angle.

So, substituting the given values, we have:
vx_initial = 56.0 m/s * cos(35.0°)
= 56.0 m/s * 0.8192
= 45.9392 m/s

Using the time of 2.90 seconds given, we can find the horizontal displacement:
Δx = 45.9392 m/s * 2.90 s
= 133.09968 m

Therefore, the horizontal distance from where the projectile was launched to where it lands is 133.09968 meters.

In summary:
The vertical distance from where the projectile was launched to where it lands is 133.1196 meters, and the horizontal distance is 133.09968 meters.

To find the x and y distances, we can use the equations of motion for projectile motion. We need to break the initial velocity into its horizontal and vertical components.

Given:
Initial speed (v0) = 56.0 m/s
Launch angle (θ) = 35.0°
Time (t) = 2.90 s

Step 1: Calculate the initial horizontal and vertical velocities:
The horizontal component of the initial velocity, vx0, can be found using the formula:
vx0 = v0 * cos(θ)

The vertical component of the initial velocity, vy0, can be found using the formula:
vy0 = v0 * sin(θ)

Using the given values:
vx0 = 56.0 m/s * cos(35.0°)
vy0 = 56.0 m/s * sin(35.0°)

Step 2: Calculate the vertical displacement:
The vertical displacement, y, can be found using the formula:
y = vy0 * t + (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Using the given values:
y = (56.0 m/s * sin(35.0°)) * 2.90 s + (1/2) * (9.8 m/s^2) * (2.90 s)^2

Step 3: Calculate the horizontal displacement:
The horizontal displacement, x, can be found using the formula:
x = vx0 * t

Using the given values:
x = (56.0 m/s * cos(35.0°)) * 2.90 s

Step 4: Solve for x and y:
Evaluate the above equations to find the values of x and y.

x = (56.0 m/s * cos(35.0°)) * 2.90 s
y = (56.0 m/s * sin(35.0°)) * 2.90 s + (1/2) * (9.8 m/s^2) * (2.90 s)^2

Calculating the above equations, you will find the values of x and y.