An attacker at the base of a castle wall 3.65 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.65 m high throws a rock straight up with initial speed 7.40 m/s from a height of 1.55 m above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is the rock

Explanation / Answer

Equations: v = v0 + at = 7.40-- 9.8 t h = h0 + v0 t + (1/2)a t² = 1.55 + 7.4 t - 4.9 t² (with t> 0) Calculate the maximum height (v = 0) v = 0 when 7.4 - 9.8 t = 0, or t = 7.4/9.8 = 0755 s At that instant, h = h(max) h(max) = 1.55 + 7.4*(0.755) - 4.9*(0.755)² h(max) = 4.34 m > hB Ans.: The rock will reach the top of the wall. b) v = 7.40-- 9.8 t h = 1.55 + 7.4 t - 4.9 t² The instant at which the rock reaches the top of the wall (h = hB) is the smaller of the positive roots of equation: 1.55 + 7.4 t - 4.9 t² = 3.65 -4.9 t² + 7.4 t - 2.10 = 0 t1 = 0.38 s t2 = 1.13 s At t1 = 0.38 s, the velocity is v1 = 7.4 - 9.8*(0.38) = 3.68 m/s Ans.: The speed of the rock at the top of the wall is 3.68 m/s c) Take as t = 0 the moment the rock was thrown from B. Initial conditions: h0 = hB = 3.65 m; v0 = -7.40 m/s Equations: h = 3.65 - 7.4 t - 4.9 t² v = -7.4 - 9.8 t (with t > 0)

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