A man stands on the roof of a building of height 13.6m and throws a rock with a

Question

A man stands on the roof of a building of height 13.6m and throws a rock with a velocity of magnitude 31.8m/s at an angle of 31.6 degrees above the horizontal. You can ignore air resistance.

A) Calculate the maximum height above the roof reached by the rock
B) Calculate the magnitude of the velocity of the rock just before it strikes the ground
C) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.


Please show work so I can follow along. Thanks!

Explanation / Answer

d = distance = -9.12 m u = initial velocity = 10.3sin51 = 8.005 m/s a = acceleration = -9.8 m/s² t = time d = ut + 1/2 at² -4.9t² + 8.005t + 9.12 = 0/ This is a quadratic equation and can be solved via the quadratic formula: At² + Bt + C = 0, then t = [-B ± v(B²-4AC)]/2A Here A = -4.9, B = 8.005 and C = 9.12. Putting in these numbers, t = [-8.005 ± v(8.005²-4*4.9*-9.12)]/2*-4.9, or t = [-8.005 ± v(242.83)]/-9.8, or t = [-8.005 ± 15.5831]/-9.8, which gives solutions of t = -0.77 or 2.41 s. Time cannot be negative...so t= 2.41 sec. (a) Velocity of the rock just before it strikes the ground V=U+at. =0+9.8*2.41/2 = 11.81 m/s (b) Horizontal distance the rock travels is =(2.41 s) times (6.482 m/s) = 15.6 m.

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