Question

if you gently push a ball off of a table top, 1.3 m above the floor. how long (in seconds) does it take the ball to reach the floor?
(a) 1.3 s
(b) 0.27 s
(c) 0.52 s
(d) 2.52 s
question 4
you drop a small stone down a stair well. it takes 2.6 s seconds to hit the ground floor. how many meters high are you above ground level?

Explanation:

Step1: Identify the kinematic - equation

The ball is in free - fall, and we use the equation $h = v_0t+\frac{1}{2}gt^2$. Since it is gently pushed off (initial vertical velocity $v_0 = 0$), the equation simplifies to $h=\frac{1}{2}gt^2$, where $h$ is the height, $g = 9.8\ m/s^2$ is the acceleration due to gravity, and $t$ is the time.

Step2: Rearrange the equation for time

From $h=\frac{1}{2}gt^2$, we can solve for $t$:
\[t=\sqrt{\frac{2h}{g}}\]

Step3: Substitute the values

Given $h = 1.3\ m$ and $g = 9.8\ m/s^2$, we have:
\[t=\sqrt{\frac{2\times1.3}{9.8}}=\sqrt{\frac{2.6}{9.8}}\approx\sqrt{0.2653}\approx0.52\ s\]

Answer:

C. 0.52 s