Tarzan, who weighs 880 N, swings from a cliff at the end of a 38.0 m vine that hangs from a high tree limb and initially makes an angle of 43.0° 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 610 N. What is the acceleration of Tarzan, as a magnitude and direction, during takeoff?

At takeoff, his velocity is zero, so there is no centripetal acceleration in the vine direction. The acceleration will be initially in the direction perpendicular to the vine, along his trajectory. The component of his weight in that direction (M g sin 43) will equal M a, so

a = g sin 43

To find the acceleration of Tarzan during takeoff, we need to first analyze the forces acting on him.

Let's break down the weight of Tarzan into its components. The vertical component of the weight is given by W_vertical = mg, where m is the mass of Tarzan and g is the acceleration due to gravity (approximately 9.8 m/s²). In this case, we are given the weight as 880 N, so we can solve for m: m = W_vertical / g = 880 N / 9.8 m/s² ≈ 89.8 kg.

The tension in the vine can be split into horizontal and vertical components. The vertical component of tension (T_vertical) counteracts the vertical component of the weight (W_vertical) to keep Tarzan from falling through the ground. The horizontal component of tension (T_horizontal) provides the accelerating force for Tarzan.

Now, let's calculate the vertical and horizontal components of the tension:
T_vertical = 610 N × cos(43.0°)
T_horizontal = 610 N × sin(43.0°)

The vertical component of the tension (T_vertical) is equal to W_vertical:
T_vertical = W_vertical
610 N × cos(43.0°) = mg

We can solve this equation for g:
g = (610 N × cos(43.0°)) / m ≈ 6.38 m/s²

Therefore, the magnitude of the acceleration during takeoff is approximately 6.38 m/s².

To determine the direction of the acceleration, we look at the sign of g. Since we assumed the positive y-axis as upward, the acceleration is in the downward direction. So, the direction of the acceleration during takeoff is vertically downward.