ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

Solve for T₁,

Now, we use the second equation to find the tension in the horizontal string,

Solve for T₂,

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
I'd say move faster, unless it's asking something else.
Answer
given,
length of the swing = 26.2 m
inclined at an angle = 28°
let, the initial height of the Tarzan be h
h = L (1 - cos θ)
a) initial velocity v₁ = 0 m/s
final velocity of Tarzan = v_f
law of conservation of energy
PE_i + KE_i = PE_f + KE_f






= 7.75 m/s
the speed tarzan at the bottom of the swing
v_f = 7.75 m/s
b)initial speed of the = 3 m/s






v_f= 11.29 m/s
One with greater mass (8kg)
Answer:
72 km/hr. ...jshdhdhddjidididdiididudd