An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 19.0 m if her initial speed is 3.80 m/s. What is the free-fall acceleration on the planet?
Solution:
To have maximum range for a given initial velocity, her launch angle must be q0 = 45o. Her range then is R = (v02sin2q 0)/g' = v02sin90o/g' = v02/g'. We have g' = v02/R = 0.96m/s2.
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To find the free-fall acceleration on the planet, you can use the horizontal range formula. The horizontal range is the maximum distance covered horizontally by a projectile.
The formula for the horizontal range (R) is:
R = (v^2 * sin(2θ)) / g
Where:
- R is the horizontal range
- v is the initial speed
- θ is the launch angle
- g is the acceleration due to gravity on the planet
In this case, the astronaut is jumping horizontally, so the launch angle is 0 degrees. Therefore, the formula simplifies to:
R = (v^2 * sin(0)) / g
R = (v^2 * 0) / g
R = 0
Since the astronaut is able to jump a distance of 19.0 m, we can conclude that the free-fall acceleration on the planet is also 0.