Answer
(a) The ball falls 44.1 meters in the first 3.0 seconds.
(b) The speed of the ball is 7.0 m/s
(c) The speed of the ball 3.0 seconds after it is released is 29.4 m/s
Work Step by Step
(a) We can find the distance the ball falls in the first $3.0~s$:
$\Delta y = \frac{1}{2}at^2$
$\Delta y = \frac{1}{2}(9.80~m/s^2)(3.0~s)^2$
$\Delta y = 44.1~m$
The ball falls 44.1 meters in the first 3.0 seconds.
(b) We can find the speed after it has traveled 2.5 m downward:
$v_f^2=v_0^2+2a\Delta y$
$v_f= \sqrt{v_0^2+2a\Delta y}$
$v_f= \sqrt{0+(2)(9.80~m/s^2)(2.5~m)}$
$v_f = 7.0~m/s$
The speed of the ball is 7.0 m/s
(c) We can find the speed of the ball 3.0 seconds after it is released:
$v_f = v_0+at$
$v_f = 0 + (9.80~m/s^2)(3.0~s)$
$v_f = 29.4~m/s$
The speed of the ball 3.0 seconds after it is released is 29.4 m/s
