Question

A particle thrown up vertically from the ground is moving up with the velocity of 30 m/s when it is at a height of 35 m above the ground. How long ago did it leave the ground?

Answer 1

Answer:

g=10m/s^2 & t=6 second

Answer 2

Answer:

The particle might have left the ground either 4.4s or 1.6s ago.

Explanation:

Given the velocity of  particle thrown vertically upwards, $u=30 m/s$

         height of the particle, $h=35 m$

        acceleration due to gravity, $-g=-10m/s^{2}$

Using the equation of motion,

                             $h=ut-\frac{1}{2} gt^{2}$

                           $35=30t-\frac{1}{2} \times10 \times t^{2}$

                           $35=30t-5 t^{2}\implies 5t^{2}-30t+35=0$

                                                    $\implies t^{2}-6t+7=0$

Using quadratic formula, $\frac{-b\pm\sqrt{b^{2} -4ac} }{2a}$,

the values of time, t can be $t=\frac{-(-6)\pm\sqrt{(-6)^{2} -4\times 1 \times 7} }{2\times 1}$

                                                $=\frac{6\pm\sqrt{36 -28} }{2}=\frac{6\pm\sqrt{8} }{2}$

                                                $=3\pm\sqrt{2} }$

                                             $t=4.4s$ or $t= 1.6s$

Therefore, the value of time is 4.4s or 1.6s