Question

A particle is projected vertically upwards from the ground level with the speed of 50 meter per second for how long will it be mire then 70m above the ground

Answer 1

According to the question, we have;

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

Now, we use the following formula;

$s = ut - \frac{gt^2}{2}$

⇒ 70 = 50t - 5t²

⇒ 5t² - 50t + 70 = 0

Now, please note that $t_1$ and $t_2$ are the two roots of equation derived from above.

So,  the smaller root = time when it crosses 70 m

Ans, the larger root = time when it comes back below 70 m

∴ $t_2 - t_1 = \sqrt{b^2 - 4ac}$ ÷ $a$ secs

⇒ $t_2 - t_1$ = $\sqrt{(50)^2 - 4 * 5 * 70}$ ÷ $a$ secs

⇒ $t_2 - t_1 =$ $2\sqrt{11}$ secs = 6.63 secs

Ans) The particle will remain in the air for 6.63 secs at a height of more than 70 m above the ground.