Question

The initial velocity of the boulder is Vo = 40 m/s and it is released at an angle ofa-55° 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 h2 - 30 m above ground level, calculate each of the following values at states 1 and 2: (a) the speed at position 1, Vi (b) the angle the velocity makes with the horizontal, 6, (c) the horizontal distance to position 1, di (d) the height of position 1, hi (e) the time of flight to position 1, ti (f) the speed at position 2, V2 (g) the angle the velocity makes with the horizontal, θ2 (h) the horizontal distance to position 2, dz (1) the time of flight to position 2,2 02 hi h2 dj d2

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