## College Physics (4th Edition)

(a) The gull should let go of the clam a horizontal distance of 7.68 meters before the rocks. (b) The speed relative to the rocks is $13.9~m/s$ (c) The speed relative to the gull is $12.5~m/s$
(a) We can find the time to fall $8.00~m$: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(8.00~m)}{9.80~m/s^2}}$ $t = 1.28~s$ We can find the horizontal distance the clam travels in this time: $x = v_x~t = (6.00~m/s)(1.28~s) = 7.68~m$ The gull should let go of the clam a horizontal distance of 7.68 meters before the rocks. (b) We can find the vertical velocity when the clam hits the rocks: $v_y = v_{0y}+gt$ $v_y = 0+(9.80~m/s^2)(1.28~s)$ $v_y = 12.5~m/s$ We can find the speed relative to the rocks: $v = \sqrt{v_x^2+v_y^2}$ $v = \sqrt{(6.00~m/s)^2+(12.5~m/s)^2}$ $v = 13.9~m/s$ The speed relative to the rocks is $13.9~m/s$ (c) Since the gull continues moving horizontally at a speed of $6.00~m/s$, the speed relative to the gull is the magnitude of the vertical velocity, which is $12.5~m/s$