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A ski jumper competing for an Olympic gold medal wants to jump a horizontal distance of 135 meters. The takeoff of the ski jump is at a height of 25 meters. What is the minimum speed the ski jumper must have at the takeoff to achieve this distance?

1) 30 m/s
2) 35 m/s
3) 40 m/s
4) 45 m/s

User Rambou
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1 Answer

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Final answer:

To solve for the ski jumper's minimum takeoff speed, the motion is treated as projectile motion. Without knowing the angle of the ski jump, the exact speed can't be determined just from the distance and height. However, it's noted that the initial speed must be greater than the horizontal component due to the vertical drop.

Step-by-step explanation:

To find the minimum speed the ski jumper must have at the takeoff to achieve a horizontal distance of 135 meters, we have to consider the motion as a projectile under gravity and ignore air resistance. The ski jump situation can be analyzed by separating the motion into horizontal and vertical components.

  • The horizontal distance covered (range) is given by the formula: Range = (v02 × sin(2θ)) / g, where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
  • The time of flight can be found from the vertical motion using the formula: Time = √((2 × height) / g), where height is the vertical distance from takeoff to landing (25 meters).
  • Since the ski jumper must cover 135 meters horizontally, we can use the time of flight to find the horizontal speed needed.

Here we assume the optimal angle for maximum range, which is 45°. However, this angle might not be practical for a ski jump. Actual ski jumpers take off at lower angles, which would require a higher speed to reach the same range.

Assuming optimal conditions and plugging in the values (g = 9.81 m/s2, height = 25 meters), we can calculate:

  1. Time of flight t: t = √((2 × 25 m) / 9.81 m/s2)
  2. Knowing t, we can then determine the horizontal velocity vx required: vx = 135 m / t
  3. This gives us the minimum speed at the horizontal component, and since the jump is at an angle, the actual speed must be greater. v0 = vx / cos(θ)

Without specific numbers for time and the angle used, we can't calculate the exact initial speed.

The correct answer from the provided options can't be determined without making assumptions about the takeoff angle. However, it's important to note that a greater initial speed than the one we calculate for the horizontal component is necessary because of the vertical component to clear the 25-meter drop. The closest speeds provided in the options may be indicative of the right range.

User VeloFX
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