Skydiver's Required Horizontal Speed for Jump

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Question:

A skydiver competing for an Olympic gold medal jumps a horizontal distance of 135 meters. The takeoff point of the skydiver is at a height of 25 meters. With what horizontal speed must she leave the jump? Group of answer choices

55 m/s

5.9 m/s

2.3 m/s

59 m/s

Answer:

To solve this problem, we need to determine the horizontal speed required for the skydiver to cover a horizontal distance of 135 meters while falling from a height of 25 meters.

First, we calculate the time it takes for the skydiver to fall 25 meters vertically. We use the formula for the time of free fall:

[ t = \sqrt{\frac{2h}{g}} ]

where:

  • ( h = 25 ) meters (height),
  • ( g = 9.8 , \text{m/s}^2 ) (acceleration due to gravity).

[ t = \sqrt{\frac{2 \times 25}{9.8}} ] [ t = \sqrt{\frac{50}{9.8}} ] [ t \approx \sqrt{5.102} ] [ t \approx 2.26 , \text{seconds} ]

Next, we use the horizontal distance formula to find the horizontal speed ( v ):

[ d = v \times t ]

where:

  • ( d = 135 ) meters (horizontal distance),
  • ( t \approx 2.26 ) seconds (time calculated above).

Solving for ( v ):

[ v = \frac{d}{t} ] [ v = \frac{135}{2.26} ] [ v \approx 59.73 , \text{m/s} ]

Therefore, the horizontal speed with which the skydiver must leave the jump is approximately 59.73 m/s. The closest answer choice is:

59 m/s