Question

A balllon is rising upward with velocity 20m/s when it is at height 100m a ball is dropped from it . find time of flight ?​

Answer 1

Answer:

Initial velocity gained by ball is u=10m/s upward and the acceleration is downward,

so first of ball will rise to a height till its velocity becomes zero this height above the *initial* position

is supposed to be H and H=

2g

u

2

=

2×10

100

=5m

then it will come back to initial point then will come back to ground.

So net distance =5+5+20=30m

Answer 2

Answer :-

6.89s

Explanation :-

Given :

Initial velocity of the ballon,u => 20m/s

Height,h => 100m

Gravitational acceleration => 10m/s^2

Note :

If a body is moving downward initially then its velocity is taken as negative and if initially body is moving upward then its velocity is taken as positive.

The acceleration is negative when going up because the speed is decreasing. The acceleration is negative when going down because it is moving in the negative direction.

To Find :

Time,t => ?

Solution :

According to the second equation of motion,

$\boxed{\sf{s=ut+\dfrac{1}{2}at^2}}$

For negative acceleration,

$\sf{}s=ut-\dfrac{1}{2}at^2$

Here,

h is same as s (displacement) and a is same as g(acceleration due to gravity)

So,

$\sf{}:\implies h=ut-\dfrac{1}{2}gt^2$

$\sf{}:\implies 100=20\times t -\dfrac{1}{2}\times 10\times t^2$

$\sf{}:\implies 100=20t -5t^2$

$\sf{}:\implies 0=-100+20t-5t^2$

$\sf{}:\implies 0=-100+20t-5t^2$

$\sf{}:\implies 5(-20+4t-t^2)=0$

$\sf{}:\implies 5\times (-20+4t-t^2)\times \dfrac{1}{5}=0\times \dfrac{1}{5}$

$\sf{}:\implies t^2-4t-20=0$

$\sf{}:\implies t^2-4t-20=0$

$\sf{}:\implies (-20-4t+t^2)=0$

$\sf{}:\implies t^2-4t-20=0$

Use quadratic formula with a=1, b=-4, c=-20.

$\boxed{\sf{t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}$

$\sf{}:\implies t= \dfrac{-(-4)\pm \sqrt{(-4)^2-4(1)(-20)}}{2(1)}$

$\sf{}:\implies t= \dfrac{4\pm \sqrt{16+80}}{2}$

$\sf{}:\implies t= \dfrac{4\pm \sqrt{96}}{2}$

$\sf{}:\implies t=\dfrac{2\pm \sqrt{96}}{1}$

$\sf{}:\implies t=2\pm \sqrt{96}$

$\sf{}:\implies t=2\pm \sqrt{\underbrace{2\times 2}\times \underbrace{2\times 2}\times 2\times 3}$

$\sf{}:\implies t=2\pm 2\times2\sqrt{2\times 3}$

$\sf{}:\implies t=2\pm 4\sqrt{6}$

2 + 4√6

⇒ 6.89

2 - 4√6

⇒ -2.89

Therefore,we get two values of t,one is 6.89 and second is -2.89.

Time is always positive,so we will not consider -2.89 as time.

Therefore,time is equal to 6.89s(Approx)