Tarzan, who weighs 820 N, swings from a cliff at the end of a 20.0 m vine that hangs from a high tree limb and initially makes an angle of 22.0° with the vertical. Assume that an x axis extends horizontally away from the cliff edge and a y axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is 760 N. Just then, what are (a) the force on him from the vine in unit-vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the x axis? What are the (e) magnitude and (f) angle of Tarzan’s acceleration just then?

i figured out everything except A)the force on tarzan from the vine

To determine the force on Tarzan from the vine, we need to break it down into its horizontal and vertical components.

(a) Force on Tarzan from the vine in unit-vector notation:
Considering upward as the positive direction for y and rightward as the positive direction for x, the force on Tarzan from the vine can be represented as:
F_vine = F_vine_x * i + F_vine_y * j

The force in the x-direction, F_vine_x, can be found using trigonometry. We can use the angle and the tension in the vine to find this component:
F_vine_x = T_vine * cos(angle)

Substituting the given values:
F_vine_x = 760 N * cos(22°)

The force in the y-direction, F_vine_y, is equal to the weight of Tarzan:
F_vine_y = weight of Tarzan = 820 N

So, the force on Tarzan from the vine in unit-vector notation is:
F_vine = 760 N * cos(22°) * i + 820 N * j

To determine the net force on Tarzan, we need to consider the gravitational force acting on him.

(b) Net force on Tarzan in unit-vector notation:
The net force on Tarzan is the vector sum of the force from the vine and the gravitational force. The gravitational force can be represented as:
F_gravity = weight of Tarzan = 820 N * j

The net force on Tarzan is given by:
Net force = F_vine + F_gravity

Substituting the values, we get:
Net force = 760 N * cos(22°) * i + (820 N + 820 N) * j

(c) Magnitude of the net force:
The magnitude of the net force can be found using the Pythagorean theorem:
Magnitude = √(Fx^2 + Fy^2)

Substituting the values, we have:
Magnitude = √((760 N * cos(22°))^2 + (820 N + 820 N)^2)

(d) Angle relative to the positive direction of the x-axis:
The angle of the net force can be found using trigonometry:
Angle = tan^(-1)(Fy/Fx)

Substituting the values, we get:
Angle = tan^(-1)((820 N + 820 N)/(760 N * cos(22°)))

(e) Magnitude of Tarzan's acceleration:
Using Newton's second law of motion, which states that the net force is equal to the mass multiplied by the acceleration:
Net force = mass * acceleration

Since we don't have the mass of Tarzan given, we cannot directly find the magnitude of his acceleration.

(f) Angle of Tarzan's acceleration:
The angle of Tarzan's acceleration can be determined using trigonometry:
Angle = tan^(-1)(ay/ax)

However, since we cannot directly find the magnitude of Tarzan's acceleration without the mass, we cannot determine its angle accurately in this case.