Question

great work!

1. now that you have ( a = \frac{x}{t^2} ), you need values for ( x ) and ( t ) to find the acceleration.

for the purposes of this activity, you will use ( x = 40 ) cm. measure how long it takes the apple to fall 40 cm once it begins falling.
show me how
( square ) s
submit answer 0 / 1 submissions used
score: 0/1

Explanation:

Step1: Recall the kinematic equation

For an object in free fall (assuming initial velocity \( u = 0\) m/s), the displacement \( x\) is given by the equation \( x=ut+\frac{1}{2}at^{2}\). Since \( u = 0\), this simplifies to \( x=\frac{1}{2}at^{2}\). We can re - arrange this formula to solve for \( t\):

$$t=\sqrt{\frac{2x}{a}}$$

But we know that for free fall near the surface of the Earth, the acceleration \( a = g\approx9.8\ m/s^{2}\) (if we assume this is a free - fall problem). First, we need to convert the distance \( x = 40\ cm=0.4\ m\) to meters.

Step2: Substitute the values into the formula

Substitute \( x = 0.4\ m\) and \( a = 9.8\ m/s^{2}\) into the formula \( t=\sqrt{\frac{2x}{a}}\):

$$t=\sqrt{\frac{2\times0.4}{9.8}}=\sqrt{\frac{0.8}{9.8}}\approx\sqrt{0.0816}\approx0.286\ s$$

(If we use \( a = 10\ m/s^{2}\) for approximation, then \( t=\sqrt{\frac{2\times0.4}{10}}=\sqrt{\frac{0.8}{10}}=\sqrt{0.08}\approx0.283\ s\))

Answer:

Approximately \( 0.29\ s\) (or more precisely around \( 0.28 - 0.29\ s\) depending on the value of \( a\) used)