A daredevil is shot out of canon at 40°above the horizontal with a speed of 25.0 m/s. If the net is located 2.00 m higher than the launch point, what is the horizontal distance to the net if it is to safely catch the daredevil?

time in air:

hf=hi+V*sinTheta*t-1/2 g t^2
hi=0, solve for time t. Use the quadratic equaiton.

Now with time t,
hordistance=V*cosTheta*t

65m?

To find the horizontal distance to the net, we need to analyze the motion of the daredevil in both the horizontal and vertical directions.

Given:
Launch speed (v₀) = 25.0 m/s
Launch angle (θ) = 40°
Vertical displacement (Δy) = 2.00 m

First, we can split the initial velocity into horizontal and vertical components:
Horizontal component: v₀ₓ = v₀ * cos(θ)
Vertical component: v₀ᵧ = v₀ * sin(θ)

Next, we need to find the time it takes for the daredevil to reach the net. We can do this by using the equation for vertical displacement (Δy) with respect to time (t) and the vertical component of velocity (v₀ᵧ):
Δy = v₀ᵧ * t + (1/2) * g * t²
Since the daredevil started and ended at the same vertical position (0 displacement in the vertical direction), we can rearrange the equation to solve for time (t).

0 = v₀ᵧ * t + (1/2) * g * t²
Rearranging the equation gives us:
(1/2) * g * t² = -v₀ᵧ * t
Dividing both sides by t (assuming t is not zero):
(1/2) * g * t = -v₀ᵧ
Simplifying, we find:
t = -2 * v₀ᵧ / g

With the time found, we can calculate the horizontal distance (Δx) traveled by the daredevil using the equation: Δx = v₀ₓ * t
Substituting in the values, we get:
Δx = v₀ * cos(θ) * t

Now, we can solve for the horizontal distance (Δx) to the net by substituting the values we have:
Δx = (25.0 m/s) * cos(40°) * [-2 * (25.0 m/s) * sin(40°) / 9.8 m/s²]

Calculating this expression will give us the horizontal distance (in meters) to the net.