I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door
)...
The door forms a right triangles that satisfies

We also have

so if you happen to know the height of the door, you can solve for
and
.
is fixed, so

We can solve for the angular velocity
:

At the point when
and
ft/s, we get

Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y =
t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
A.lucid because it makes more sense yo the answer how about you do it yourself
We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2