Question

A particle is projected vertically upwards from ground level with a speed of $50m/s$. Fir how long will it be more than $70m$ above the ground?

• A. $2.63sec$
• B. $4.63sec$
• C. $6.63sec$
• D. $8.63sec$

Answer 1

The correct option is C $6.63sec$

$s=70m,u=50,a=-10$

using:$s=ut-\frac{g{t}^2}{2}$

$70=50t-5t^2$

$5t^2-50t+70=0$

$t_1\&t_2$ are two roots of above equation

then smaller root is :time when it crosses 70m

larger root is : time when it comes back below 70m

$t_2-t_1=\frac{\sqrt{b^2-4ac}}{a}$

$=\frac{\sqrt{\left(\right(-50{)}^2-4\times 5\times 70}}{5}$

$=2\sqrt{11}$

$=6.63sec$