Answer:
2.03873 s
70.38735 m
4.07747 seconds
5.82688 seconds
27.37482 m from the ground
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
time needed for the ball to reach its maximum height is 2.03873 s
The time taken to go up and the time taken to reach the point from where it was thrown is the same.
So, time needed for the ball to return to the height from which it was thrown is 2.03873+2.03873 = 4.07747 seconds
The maximum height the ball will reach is 50+20.38735 = 70.38735 m
Time needed to reach the ground is 2.03873+3.78815 = 5.82688 seconds
The time from the maximum height that is required is 5-2.03873 = 2.96127 seconds
The ball will be 70.38735-43.01253 = 27.37482 m from the ground
Lets use the expression
F = k*x
Where k is the spring constant [N/m]
And x the distance from the equilibrium position.
This force is equal to the force due to the acceleration of gravity
F = k*x = m*g
F = 525[N/m]* 0.30 [m] = 157.5 [N]
157.5 [N] = m *10[m/s**2] ...................> m = 15.75 kg
Answer:
v = 10 m/s
Explanation:
Given,
Length of the vine, l = 36 m
Angle of inclination = 31.0◦ with the vertical
acceleration due to gravity = 9.81 m/s²
Using Conservation of energy
KE = PE
v = 10 m/s
Hence, the speed of the swing is equal to 10 m/s