A projectile of mass 0.500 kg isshot straight up with an initial speed of 22.0 m

Question

A projectile of mass 0.500 kg isshot straight up with an initial speed of 22.0 m/s. (a) How high would it go if there were no air friction?
m

(b) If the projectile rises to a maximum height ofonly 17.8 m, determine themagnitude of the average force due to air resistance.
N
(a) How high would it go if there were no air friction?
m

(b) If the projectile rises to a maximum height ofonly 17.8 m, determine themagnitude of the average force due to air resistance.
N

Explanation / Answer

(a) h = u2/ 2g         = 22 * 22 / 2 * 9.8         = 24.69 m (b) PE = mgh = (0.5)(9.8)(17.8) = 87.22J.      KE = 1/2mv2 =1/2(0.5)(22)2 =121J.     So the energy (or work) that was lost tofriction was w = 121 - 87.22 = 33.78J.     Now we can find the force, soW=FdCos                                                     33.78 = F * 17.8 * cos180                                          ==>          F = -1.897 N Magnitude is 1.897 N

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