Answer
(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 $12.5~m/s$
Work Step by Step
(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 $12.5~m/s$
