Question

Question: a particle is projected in vertically upward direction such that it crossed height of 80 m twice in its Journey with time interval being 6 second. Then find the max height
attained by the particle
​

Answer 1

Answer:

The maximum height attained by the body is 125m

Explanation:

Here, u = u m/s ; g = −10m/s  and h = 80m.

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

$=> 80 = ut-5t^2\\=>5t^2-ut+80=0$

Where the difference of 2 roots of 't' in this equation gives time interval 6s.

So, the 2 roots of 't' are

$t=\frac{u+\sqrt{u^2-1600} }{10} \\\\t=\frac{u-\sqrt{u^2-1600} }{10}$

The difference between them is

$6=\frac{u+\sqrt{u^2-1600} }{10} -\frac{u-\sqrt{u^2-1600} }{10} \\\\6=\frac{u+\sqrt{u^2-1600}-u+ \sqrt{u^2-1600}}{10}\\\\6=\frac{2\sqrt{u^2-1600} }{10} \\\\\sqrt{u^2-1600} =30\\\\u^2-1600=900\\\u=\sqrt{2500} \\\\u=+50 or-50$

Now we'll find the maximum height

$=>H=\frac{u^2}{2g}$

$=>H=\frac{50^2}{20} \\\\\=>H=\frac{2500}{20} \\\\=>H=125m$

Therefore, the maximum height reached by the particle is 125m

Thankyou :)