Question

A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall

Answer 1

Answer:

The object fell from about 38.14 meters

Explanation:

We can use the formula for displacement under accelerated motion due to gravity to find the velocity of the object had 30 m before hitting the ground :

$y_f-y_i=v_i * \,t-\frac{9.8}{2} \,t^2\\0-30=v_i \,(1.5)-4.9\,(1.5)^2\\-18.975=v_i\,(1.5)\\v_i=-12.65\,\,\frac{m}{s}$

Now, knowing this velocity, we can find the time it took to fall from the initial position to 30 m before hitting the ground:

$v_f=v_i-g\,*\,t\\-12.65=0-g \,*\,t\\t=\frac{12.65}{g} \\t\approx1.29\,\,s$

And now we can find what is the total distance covered in 1.5 s plus 1.29 seconds for this free falling object:

$y_f-y_i=v_i * \,t-\frac{9.8}{2} \,t^2\\0-y_i=0-\frac{9.8}{2} \,(2.79)^2\\-y_i=-38.14\\y_i=38.14\,\,m$