Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity

m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting

The center of mass of the two-ball system is 7.05 m above ground.
Answer:
a) 4.45 m/s
b) 0.9 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²

a) The vertical speed when the player leaves the ground is 4.45 m/s

Time taken to reach the maximum height is 0.45 seconds

Time taken to reach the ground from the maximum height is 0.45 seconds
b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds
Simply, apply the formula

and insert the values of m = mass, v = velocity and E = Energy.
The result will be

, m = 1 kg
Answer:
40 N
Explanation:
F=ma where F is the applied force, m is the mass of object and a is the acceleration.
Since there is no friction, substituting 20 Kg for m and 2 m/s squared for a then we obtain
F=20*2=40 N