King Arthur\'s knights used a catapult to launch a rock from theirvantage point

Question

King Arthur's knights used a catapult to launch a rock from theirvantage point on the top of a castle wall, 12 m above the moat. The rock is launched at a speed of 30 m/s and an angle of 29degrees above the horizontal. How far from the castle walldoes the launched rock hit the ground? With this problem is motion both in the x and y directions? I'm not sure which variables I have, and I'm confused onwhich kinematic equation to use to solve this. Can anyonehelp? With this problem is motion both in the x and y directions? I'm not sure which variables I have, and I'm confused onwhich kinematic equation to use to solve this. Can anyonehelp?

Explanation / Answer

= 29 degrees v initial = 30m/s y = y final - y initial = 0 - 12 m = -12m x = ? a = -9.8m/s^2 you use the equation to find the time y = (vsin )t + 0.5at^2 and then plug it into this equation to find x x = (vcos)t y = (vsin )t + 0.5at^2 -12 = (30sin 29)t + (0.5)(-9.8)t^2 -12 = 14.544 t - 4.9 t^2 0 = 4.9t^2 -14.544t -12 t = 3.6408      t =-0.6726       time can not benegative so we will use t = 3.6408 x = (vcos)t       = (30 cos 29) (3.6408)       = 26.239(3.6408)       = 95.529 m

Related Questions

Navigate

• Browse (All)
• Browse K
• Subjects

---

---

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