Question

An object is thrown vertically up from the ground passes the height 5m twicd in an interval of 10s.What is the time of flight

Answer 1

Given :-

• An object thrown vertically upwards passes 5m height twice at an interval of 10s .

To Find :-

• The time of flight .

Solution :-

Considering the motion symmetrical ( neglecting drag ) , it's velocity at same height will be same when it goes up and down , say v . Hence time of going up and down will be same .

• Let's take that it took t time to go from A to C and from D to B .
• At maximum height velocity will be 0 .

From first equation of motion ,

$\implies\sf v = u + at \\\\\implies\sf 0 = u_c - 10(5) m/s \\\\\implies \sf u_C = 50m/s$

From second equation of motion ,

$\implies \sf s = ut +\dfrac{1}{2}at^2\\\\\sf\implies s = (50)(5) m-\dfrac{1}{2}(10)(5^2)m \\\\\sf\implies s = 250 m- 125m \\\\\sf\implies s = 125m$

Hence , total height , that is maximum height ,

$\sf\implies h_{max}= 125m + 5m \\\\\sf\implies h_{max}= 130m$

From Third equation of motion ,

$\sf\implies 2as = v^2-u^2\\\\\sf\implies 2(-10)(130) = 0^2- (u)^2 \\\\\sf\implies u^2 = 2600 \\\\\sf\implies u =\sqrt{(26)(100)}m/s \\\\\sf\implies u = 10\sqrt{26} m/s$

As , we know that time of flight is ,

$\implies\sf T_f =\dfrac{2u}{g}\\\\\sf\implies T_f = \dfrac{2(10\sqrt{26}}{10} s \\\\\sf\implies \boxed{\red{\sf Time_{flight}= 2\sqrt{26} s }}$