When an object falls or is dropped from rest it's initial velocity is zero.
Using the equations for a motion in straight line. I can find the time it takes to reach 3.0 m down (half way).
x = vt - 4.9t²
-3 = 0 - 4.9t²
-3/-4.9 = t²
0.6122 = t²
0.7825 sec = t
v = v - gt
v = 0 - 9.8(0.7825)
v = -7.67 m/s
the negative denotes downward direction.
You could also solve the problem using potential and kinetic energy.
Since it starts with maximum PE and gets converted to KE when it hits the ground. mgh = mv²/2
mass cancels, use 3 meters for the halfway distance
-9.8(-3) = v²/2
29.4 * 2 = v²
√(58.8) = 7.67 m/s downwards
Explanation:
1.
We use the equation
h =
, where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be

= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
Answer:

Explanation:
From the question we are told that
Angle of cable 2 
Weight of sculpture 
Generally the Tension from cable 2
is mathematically given by



Generally the Tension from Cable 1
is mathematically given by


