Question

I have been working on this for 2 days and I cannot get the correct answer. Any help would be greatly appreciated.
A child running along level ground at the top of a 30-ft high vertical cliff at a speed of 15 ft/s, throws a rock over the cliff into the sea below. Suppose the child's arm is 3 ft above the ground and her arm speed is 25 ft/s. If the rock is release 10 ft from the edge of the cliff at an angle of 30 degrees,
1. How long does it take for the rock to hit the water?
2. How far from the base of the cliff does it hit?

Answer 1

The picture shows the question drawn out with values as follows:\[h = 30+3 =33\]\[u=25\]\[\theta = 30^o\]1. Time of flight?\[2s=2ut+at^2\]\[2 \times -33 = 2 \times 25 \sin (30) t - 9.8 t^2\]\[-9.8t^2+50 \sin (30) t +66 =0\]\[t \approx 4.167\] 2. How far? \[2s = 2ut +at^2\]\[a_x=0\]\[s = 25 \cos (30) \times t_{total}\]\[s \approx 90.222\]

Answer 2

The s, by the way, is the displacement, so How far from the base of the cliff is 90.222-10=80.222