Question

A ball rolls off a table with a horizontal velocity of 4 m/s. If it takes 0.5 seconds for the ball to reach the floor, how high above the floor kd tje tabletop? (Use g = 10 m/s^2)

Show work please :)​

Answer 1

Answer: 10m

Explanation: Workings

h = ½gt²

h = ½ × 10 × 4²

h = 5 × 2

h = 10 m

Hope this helps.

Answer 2

Answer:

1.25 m

Explanation:

Constant Acceleration Equations (SUVAT)

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm} s = displacement in m\\\\ u = initial velocity in ms ^{-1} \\\\ v = final velocity in ms ^{-1} \\\\ a = acceleration in ms ^{-2} \\\\ t = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Consider the horizontal and vertical motion of the ball separately.

As the ball rolls off the table with a horizontal velocity only, the vertical component of its initial velocity is zero.

Acceleration due to gravity = 10 ms⁻²

As we need to find the vertical displacement of the ball, resolve vertically, taking ↓ as positive:

$u=0 \quad a=10 \quad t=0.5$

$\begin{aligned}\textsf{Using} \quad s&=ut+\dfrac{1}{2}at^2\\\\s&=(0)(0.5)+\dfrac{1}{2}(10)(0.5)^2\\s&=0+\dfrac{1}{2}(10)(0.25)\\s&=(5)(0.25)\\ \implies s&=1.25 \; \sf m\end{aligned}$

Therefore, the vertical displacement of the ball is 1.25 m, and so the table is 1.25 m above the floor.