Question

example 1
a 50 kg object falls 490.5 m. what is the speed of the object just before impact with the ground?

Explanation:

Step1: Select kinematic equation

We use the kinematic equation for free fall (initial velocity $u=0$, acceleration $a=g=9.8\ \text{m/s}^2$, displacement $s=490.5\ \text{m}$):
$$v^2 = u^2 + 2as$$

Step2: Substitute known values

Substitute $u=0$, $a=9.8\ \text{m/s}^2$, $s=490.5\ \text{m}$:
$$v^2 = 0^2 + 2\times9.8\times490.5$$

Step3: Calculate $v^2$

Compute the right-hand side:
$$v^2 = 2\times9.8\times490.5 = 9611.4$$

Step4: Solve for final velocity $v$

Take the square root of $v^2$:
$$v = \sqrt{9611.4}$$

Answer:

$98.04\ \text{m/s}$ (or approximately $98\ \text{m/s}$)