Asked by Anees
A projectile is launched with a speed Vo at an angle 45° with horizontal find ratio of its minimum and maximum K.E ?
Answers
Answered by
R_scott
neglecting air resistance ... max K.E. is at launch ... 1/2 * m * Vo^2
min K.E. is at max height
... zero vertical velocity ... horizontal velocity equals ... Vo cos(45º)
... 1/2 * m * (Vo * √2 / 2)^2 = m * Vo^2 / 4
min K.E. is at max height
... zero vertical velocity ... horizontal velocity equals ... Vo cos(45º)
... 1/2 * m * (Vo * √2 / 2)^2 = m * Vo^2 / 4
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