A ball is dropped from the top of a tall building. It took 3.0 secs for the ball to hit the ground. The acceleration of gravity is g = 10 m/sec^2. The initial velocity was 0 m/sec. What was the ball's downward velocity?

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Answer 1 · 2018-02-07T15:45:20

Velocity of ball when it hits the ground is $30 m/s$ $"30 m/s"$

$v = u + a t$ $v = u + at$

Use above equation of motion.

$v = 0 m/s + ({10 m/s}^{2} \times 3.0 s) = 30 m/s$ $v = "0 m/s" + ("10 m/s"^2 × "3.0 s") = "30 m/s"$

Answer 2 · 2018-02-07T15:47:32

$30 \frac{m}{s}$ $30m/s$

Use one of the kinematic equations:

$v = v_{0} + a t$ $v=v_0+at$ .

Where $v$ $v$ is the final velocity, $v_{0}$ $v_0$ the initial velocity, $a$ $a$ the acceleration and $t$ $t$ the time.

Here,
$v_{0} = 0 \frac{m}{s}$ $v_0=0m/s$
$a = 10 \frac{m}{s^{2}}$ $a=10m/s^2$
$t = 3 s$ $t=3s$

Input:

$v = 0 + 10 \cdot 3$ $v=0+10*3$

$v = 30 \frac{m}{s}$ $v=30m/s$