Question

An aeroplane is moving with a velocity u. It drops a packet from height h. The time t taken by the packet in reaching the ground will be 

(answer should come √2h /√g

Answer 1

Vertical distance traveled by packet$= h = u t + \frac{1}{2} g t^{2} \\ \\ h = 0 t + \frac{1}{2} g t^{2} \\ \\ t = \sqrt{\frac{2h}{g}}$

Answer 2

Given,

The velocity of the flying aeroplane is = u

The packet was dropped from the height of = h

To find,

The time taken by the packet to reach the ground.

Solution,

As,the plane was flying in the horizontal direction and the packet was dropped in vertical direction,so the x component of the initial velocity of the packet will be "u" and the y component of the initial velocity of the packet will be zero.

Now,the packet falls in the direction of the y component so we will assume that the packet has an initial velocity of zero.

So,to calculate the time taken by the packet to cover "h" distance we need to apply the following mathematical formula,

h = ut + 1/2 gt² [ h = height, u = initial velocity, g = gravitational acceleration, t = time.]

h = 0×t + 1/2 gt²

h = 1/2 gt²

2h = gt²

gt² = 2h

t² = 2h/g

t = ✓2h/✓g

Hence,the time taken by the packet to reach the ground is ✓2h/✓g unit