Historical researchers are attempting to reenact the siege of a particular mount

Question

Historical researchers are attempting to reenact the siege of a particular mountain fortress by launching a spherical boulder, m = 100 kg, from a medieval trebuchet. The initial velocity of the boulder is V_0 = 38 m/s and it is released at an angle of theta_0 = 55 degree above the horizontal as shown in the diagram below. Assume that the boulder is released at ground level at position 0. The projectile reaches its maximum height at position 1 and lands at its target at position 2. Given that the desired target is at a height of h_2 = 25 m above ground level, calculate each of the following values at states 1 and 2: (a) the speed at position 1, V_1 (b) the angle the velocity makes with the horizontal. theta_1 (c) the horizontal distance to position 1, d_1 (d) the height of position 1, h_1 (e) the time of flight to position 1, t_1 (f) the speed at position 2, V_2 (g) the angle the velocity makes with the horizontal, theta_2 (h) the horizontal distance to position 2, d_2 (i) the time of flight to position 2, t_2

Explanation / Answer

(A)at highest position, there will be only horizontal component.

vertical component of velocity = 0

V1 = 38 cos55 = 21.8 m/s

(B) theta1 = 0

(C) at 1: vy = 0

v0y = 38 sin55 = 31.13 m/s

Applying vf = vi + a t

0= 31.13 - 9.8t

t = 3.18 sec

d1 = vx t = 21.8 x 3.18 = 69.24 m

(d) h = voy t + ay t^2 /2

= 31.13 x 3.18 - 9.8(3.18^2)/2

= 49.4 m

(e) t1 = 3.18 sec

(f) y = v0y t + ay t^2 /2

25 = 31.13t - 4.9t^2

4.9t^2 - 31.13t + 25 = 0

t = 5.41 sec

vy = 31.13 - (9.8 x 5.41) = -21.9 m/s

vx = 21.8 m/s

v = sqrt(vx^2 + vy^2) = 31 m/s

(g) theta2 = - tan^-1(21.8/21.8) = - 45 deg

45 deg below the horizontal.

(h) d2 = vx t = 21.8x 5.41 = 118 m

(i) t2 = 5.41 sec

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