a 500 Newton ballerina did 250 joules of work to lift herself upwars through the air. She landed a total of 2.5 meters to the left after completing her jump. How high did she jump?
2 answers
You are given enough information to solve for the takeoff velocity, "hang time" and "launch angle", but I could find no real solution for the hang time, using your inputs.
Her mass is M = Weight/g = 51.0 kg
Her initial velocity, from her initial kinetic energy E (equal to work done), is
Vo = sqrt(2 E/M) = 3.13 m/s
The horizontal distance X that she travels is related to her "takeoff angle" A by the equation
X = 2.5 = (Vo^2/g)*sin(2A)
which leads to
2.5 = 1.00 sin (2A)
There is no angle 2A (or A) that satifies this equation. She does not have enough initial kinetic energy to jump that far. You may have copied something wrong
Her initial velocity, from her initial kinetic energy E (equal to work done), is
Vo = sqrt(2 E/M) = 3.13 m/s
The horizontal distance X that she travels is related to her "takeoff angle" A by the equation
X = 2.5 = (Vo^2/g)*sin(2A)
which leads to
2.5 = 1.00 sin (2A)
There is no angle 2A (or A) that satifies this equation. She does not have enough initial kinetic energy to jump that far. You may have copied something wrong
1 answer