Question

i. Solve the following problem on a clean sheet of paper. A cannon shoots a cannonball from the ground level (ignore the height of the cannon) towards a cliff of height h = 170 m. The cannonball is launched with an initial velocity of 110 m/s at an angle of 64° above the horizontal. Neglect air resistance. Assume the cannonball hits the cliff as it descends (on its way down) exactly at the edge, as shown. Landing point to a. Determine the maximum height above the ground reached by the cannonball. b. What speed will the cannonball have at the maximum height? c. Determine the time it takes the cannonball to hit the cliff. d. How far away from the base of the cliff is the cannon? e. What will be the speed of the ball as it hits the cliff? f. What is the direction of the velocity of the ball as it hits the cliff?

please help am doubting my analysis

Answer 1

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a) At max height, V_y = 0 so, v^2_y, f = v^2_y, c - 2gH or, O = [(110 m/s) (sin 64 degree)]^2 - 2gH or, H = 498.71 m b) at max height v_y = 0 so, v = v_x, f = v_x, i = v_0 cos theta = (110 m/s) (cos 64 degree) = 48.22 m/s c) h = v_i, y t - 1/2 gt^2 or, 170 m = (110 m/s) (sin 64 degree) t - 1/2 (9.8 m/s^2) t^2 or, t = 18.28 s d) x = v_x t = (110 m/s) (cos 64 degree) t = 881.5m e) v_x, f = v_x, i = (110 m/s) (cos 64 degree) = 48.22 m/s v_y, f = v_y i - ft = (110 m/s) (sin 64 degree) - g(18.285) = -80.27 m/s = (110 m/s) (sin 64 degree) - g(18.285) = -80.27 m/s so, v = squareroot v^2_x, f + v^2_y,. f = 93.64 m/s