A dart is thrown at a dartboard 3.90 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.555 s

At what angle relative to the floor was the dart thrown?

0 degrees?

3.9 = vo cosθ (.555)

½ g (5552 ) = vo sinθ
Divide eq 2 by eq 1 to clear vo do a tan-1 to solve

To find the angle at which the dart was thrown, we need to use the formula for projectile motion. In this case, we can use the horizontal and vertical components of the motion.

First, let's find the initial vertical velocity (Vy) of the dart when it was released. We know that the dart hit the center of the dartboard, which means that the vertical displacement (Δy) is zero. The time of flight (t) is given as 0.555 s. We can use the formula:

Δy = Vy * t + (1/2) * g * t^2

Since Δy is zero, we can rearrange the formula to solve for Vy:

Vy * t + (1/2) * g * t^2 = 0

(1/2) * g * t^2 = -Vy * t

Vy = -(1/2) * g * t

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

Next, let's find the horizontal velocity (Vx) of the dart. We need to use the formula:

Δx = Vx * t

where Δx is the horizontal displacement and t is the time of flight. We know that Δx is 3.90 m, so we can rearrange the formula to solve for Vx:

Vx = Δx / t

Now that we have the vertical and horizontal velocities, we can find the angle (θ) at which the dart was thrown. We can use the following formula:

tan(θ) = Vy / Vx

To find θ, we can take the inverse tangent (arctan) of the ratio of Vy to Vx:

θ = arctan(Vy / Vx)

Plug in the values we calculated to find θ.