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A skydiver jumped out of a plane at 310 m high. If the landing spot is 176 m away, how fast was the skydiver initially moving?

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Final answer:

The skydiver was initially moving at a speed of about 16.8 m/s in the upward direction.

Step-by-step explanation:

To calculate the initial velocity of the skydiver, we can use the equations of motion. The skydiver jumped from a height of 310 m and landed 176 m away. We can assume that the only force acting on the skydiver is gravity, which would cause him to accelerate downward.

Using the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time, we can solve for t:

310 = (1/2)(9.8)t^2

t^2 = 310/(0.5 * 9.8) = 63.265

t = sqrt(63.265) = 7.95 s

Now, we can use the equation s = ut + (1/2)gt^2, where s is the distance, u is the initial velocity, g is the acceleration due to gravity, and t is the time to solve for u:

176 = u * 7.95 + (1/2)(9.8)(7.95)^2

176 = u * 7.95 + 0.5 * 9.8 * 63.025

176 = u * 7.95 + 309.54

176 - 309.54 = u * 7.95

-133.54 = u * 7.95

u = -133.54/7.95

u = -16.8 m/s

The skydiver was initially moving at a speed of about 16.8 m/s in the upward direction.

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