Question

A cup falls off of a table from a height of 0.75m. What is the impact speed of the cup?

Answer 1

Answer:

3.822 m/s

Explanation:

You can solve this by using the kinematic equations:

$v_{f} = v_{i}+gt$

$d = v_it\times\dfrac{1}{2}gt^{2}$

Where:

Vf = final velocity (impact velocity)

Vi = initial velocity

g = acceleration due to gravity

d = distance traveled

t = time

Acceleration due to gravity on Earth is constant. Gravity accelerates objects towards the ground at 9.8m/s².

Initial velocity is always at 0 m/s in free fall. Let's see what we have as our given:

d = 0.75m

Vi = 0 m/s

g = 9.8 m/s²

Look at our equation for the impact velocity (Vf) with our current given plugged in:

$v_{f}=v_{i}+gt$

$v_{f}=0m/s+(9.8m/s^{2})(t)$

We still do not have time. That is where the second equation comes in. We plug in our values again in the secon equation and derive time:

$d=v_{i}t+\dfrac{1}{2}gt^{2}\\\\0.75m=0m/s(t)+\dfrac{1}{2}(9.8m/s^{2})(t^{2})\\\\0.75m = 0+(4.9m/s^{2})(t^{2})\\\\\dfrac{0.75m}{4.9m/s^{2}}=t^{2}\\\\\sqrt{ 0.15s^{2}}=\sqrt{t^{2}}\\\\0.39s=t$

So our time is 0.39s. Now we use this in our first equation:

$v_{f}=0m/s+(9.8m/s^{2})(t)$

$v_{f}=(9.8m/s^{2}(0.39s)$

$v_{f}=3.822m/s$