Skydiver's Required Horizontal Speed for Jump

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

To find the horizontal speed at which the skydiver must leave the jump, we can use the principles of projectile motion. The horizontal distance traveled (range) and the vertical drop (height) can be used to determine the required horizontal speed.

1. Calculate the time of flight: The time it takes for the skydiver to fall from a height of 25 meters can be calculated using the formula for free fall:

[ h = \frac{1}{2} g t^2 ]

where:

• ( h ) is the height (25 m),
• ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )),
• ( t ) is the time in seconds.

Rearranging the formula to solve for ( t ):

[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 25 , \text{m}}{9.81 , \text{m/s}^2}} \approx \sqrt{\frac{50}{9.81}} \approx \sqrt{5.1} \approx 2.26 , \text{s} ]

2. Calculate the horizontal speed: The horizontal speed ( v ) can be calculated using the formula:

[ \text{horizontal distance} = \text{horizontal speed} \times \text{time} ]

Rearranging to solve for horizontal speed:

[ v = \frac{\text{horizontal distance}}{t} = \frac{135 , \text{m}}{2.26 , \text{s}} \approx 59.74 , \text{m/s} ]

Rounding to two significant figures, the horizontal speed is approximately ( 59 , \text{m/s} ).

Thus, the correct answer is:

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