a balloon is moving upwards with speed of 5m per second .when it is at height 98 m.a packet dropped from it .what velocity of packet when it strikes the ground

Final energy=initial energy

1/2 m vfinal^2=mg(98)+1/2 m 5^2

solve for vfinal

11m/s

To find the velocity of the packet when it strikes the ground, we need to consider the motion of the balloon as well as the free-fall motion of the packet.

Let's break down the problem:

1. The balloon is moving upwards with a speed of 5 m/s. This means the initial velocity of the packet is also 5 m/s upward relative to the balloon.

2. The packet starts falling freely from the moment it is released. Gravity acts on the packet, causing it to accelerate downward at a rate of 9.8 m/s² (assuming no air resistance).

3. The height of the packet when it is released is 98 m.

To solve for the final velocity of the packet when it strikes the ground, we can use the equation of motion:

Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))

Since the packet starts from rest (relative to the ground) and accelerates downward under gravity, we can rearrange the equation as follows:

v² = u² + 2ad

Where:
u = 5 m/s (initial velocity relative to the balloon)
a = 9.8 m/s² (acceleration due to gravity)
d = 98 m (height of the packet when it is released)

Substituting the given values into the equation:

v² = (5 m/s)² + 2(9.8 m/s²)(98 m)

v² = 25 m²/s² + 1960 m²/s²

v² = 1985 m²/s²

Taking the square root of both sides, we find:

v ≈ 44.5 m/s

Therefore, the velocity of the packet when it strikes the ground is approximately 44.5 m/s downward.