Answer:
A
Explanation:
take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time)
It can be disproved with <span>one reproducible experiment.</span>
<span>We never really used the acronym "IMA", or ideal mechanical advantage, but I'm assuming you are trying to increase the leverage and ease the effort. If so, the answer is false. You want larger movement on the effort side, and smaller movement on the resistant side of the fulcrum.</span>
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