A man stands on the roof of a 10.0 -m tall building andthrows a rock with a velo

Question

A man stands on the roof of a 10.0 -m tall building andthrows a rock with a velocity of magnitude 24.0 m/s atan angle of 38.0 above the horizontal. You can ignore airresistance. Calculate the magnitude of the velocity of the rock justbefore it strikes the ground. A man stands on the roof of a 10.0 -m tall building andthrows a rock with a velocity of magnitude 24.0 m/s atan angle of 38.0 above the horizontal. You can ignore airresistance. Calculate the magnitude of the velocity of the rock justbefore it strikes the ground. A man stands on the roof of a 10.0 -m tall building andthrows a rock with a velocity of magnitude 24.0 m/s atan angle of 38.0 above the horizontal. You can ignore airresistance. Calculate the magnitude of the velocity of the rock justbefore it strikes the ground.

Explanation / Answer

The time (t) taken by the rock to touch the ground is                        yf - yi = (vsin)t -(1/2)gt2                         0 - 10.0m = (24.0m/s)(sin38o)t -(1/2)(9.8m/s2)t2                            -10.0 = 14.8t - 4.9t2    4.9t2 - 14.8t- 10.0 = 0       t = 3.59s Now horizontal component of speed of the rock when it touches theground is       vx =(24.0m/s)(cos38o) = 18.91 m/s Vertical component of the speed of the rock when it touches theground is                               vy = vsin - gt                                   = (24.0m/s)(sin38o) - (9.8m/s2)(3.59s)                                   = -20.41m/s Therefore the magnitude of the speed of the rock when it touchesthe ground is                           v = [vx2 +vy2]                              = 27.8m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.