Question

A weight is dropped from the top of a building 135m high. How fast is the weight moving just before it hits the ground?

Answer 1

There are at least two methods to solve this problem, one with kinematics and the other with a combination of kinematics and energy conservation. These are just the two methods that come to mind. I will give an explanation of an easy method because it will help student to understand quickly

Complete step by step solution:
This is a problem on the projectile motion that can be solved using kinematics. Because the weight is dropped from rest, we know that its initial velocity is $0$ that is ${v_i} = 0$. In the question they are also given that it is dropped from a height $135m$ that is ${y_i} = 100m$ because the weight hits the ground we can take its final velocity as $0$ that is ${y_f} = 0$ here object experience free fall so we know that the acceleration is $- g$ that is $- 9.8m/{s^2}$
Now we can use the kinematic equation to solve the final velocity ${v_f}$
$v_f^2 = v_i^2 + 2{a_y}\Delta y$
Whereas $\Delta y = {y_f} - {y_i}$ as we determined above ${v_f} = 0$ by using that we have to solve for ${v_f}$
${vf} = \sqrt {2{a_y}\Delta y}$
Now substitute our known values in the above equation
${v_f} = \sqrt {2( - 9.8m/{s^2})(0 - 135)}$
Therefore after solving the above equation we get
${v_f} = 44.3m/s$

Therefore the final velocity is $44.3m/s$

Note: as we know the problem is dependent upon the projectile motion let us see what is meant by projectile motion.
Projectile motion is defined as a form of a motion by an object or particle that is projected near the Earth's surface and moves along the curved path under the action of gravity only. This curved path should be a parabola, but may also be a line in the special case when it is thrown directly upwards