An object is dropped from certain height (from point P). It crosses 2 points A and B in interval of 2 seconds such that AB = 80 m. Find distance AP in meters. (Take g = 10 m/s²)

The correct answer is 45.

Step-by-step Explanation:

Let the object be dropped from point P with an initial velocity of zero:
$u_P=0\text{m/s}$

As shown in the image, the object falls vertically from point P, passing through point A and then point B. Let the time taken to travel from P to A be $t_1$. The velocity of the object at point A is:
$v_A=u_P+g{t}_1=g{t}_1$

We are given that the object travels from point A to point B in a time interval of:
$t=2\text{seconds}$
and the distance between A and B is:
$\mathrm{AB}=80\text{m}$

Using the second equation of motion for the path AB:
$s_{\mathrm{AB}}=v_{A}t+\frac{1}{2}g{t}^2$

Substituting the given values into the equation (with $g=10{\text{m/s}}^2$):
$80=v_A\left(2\right)+\frac{1}{2}\left(10\right){\left(2\right)}^2$
$80=2v_A+5\left(4\right)$
$80=2v_A+20$
$2v_A=60$
$v_A=30\text{m/s}$

Now, we can find the time $t_1$ taken to travel from P to A using the velocity at A:
$v_A=g{t}_1$
$30=10t_1$
$t_1=3\text{seconds}$

The distance AP traveled during this time interval $t_1$ is given by:
$s_{\mathrm{AP}}=\frac{1}{2}g{t}_1^2$
$s_{\mathrm{AP}}=\frac{1}{2}\left(10\right){\left(3\right)}^2$
$s_{\mathrm{AP}}=5\times 9=45\text{meters}$

Thus, the distance AP is 45 meters.