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
Maximum torque.
T=F.r
r is measured from point of turning, which is the hinge. r is maximum when the handle is at the free edge.
To solve this problem we will apply the concepts related to potential gravitational energy. This is defined as the product between mass, acceleration and change in height and can be expressed as,
Here,
m = Mass
g = Gravitational acceleration
= Height
Replacing with our values we have,
Therefore the change in gravitational potential energy is 883J.
