An aeroplane is flying at height h with horizontal velocity u. The velocity of a dropped packet on reaching the ground will be
Understanding the Velocity of a Dropped Packet from an Aeroplane
When a packet is dropped from an aeroplane flying horizontally, it becomes a projectile. The motion of the packet can be analyzed by considering its horizontal and vertical components separately. We assume that air resistance is negligible.
Analyzing Horizontal Velocity
Initially, the aeroplane has a horizontal velocity \(u\). Since the packet is dropped from the aeroplane, it also starts with the same horizontal velocity \(u\). Once dropped, there is no horizontal force acting on the packet (assuming no air resistance). Therefore, the horizontal velocity of the packet remains constant throughout its flight, equal to the initial horizontal velocity \(u\). Let's denote the final horizontal velocity on reaching the ground as \(v_x\). So, \(v_x = u\).
Analyzing Vertical Velocity
The packet is dropped from a height \(h\). Initially, the packet only has a horizontal velocity, so its initial vertical velocity is zero. As the packet falls, it is under the influence of gravity, which causes a constant downward acceleration \(g\). We can use the equations of motion to find the final vertical velocity just before the packet hits the ground. Let's denote the initial vertical velocity as \(v_{0y} = 0\) and the final vertical velocity as \(v_y\). The vertical displacement is \(h\).
Using the kinematic equation \(v^2 = u^2 + 2as\) for the vertical motion:
\(v_y^2 = v_{0y}^2 + 2gh\)
Since \(v_{0y} = 0\), we get:
\(v_y^2 = 0^2 + 2gh\)
\(v_y^2 = 2gh\)
So, the magnitude of the vertical velocity on reaching the ground is \(v_y = \sqrt{2gh}\).
Calculating the Final Velocity Magnitude on Reaching the Ground
The velocity of the dropped packet on reaching the ground is the vector sum of its final horizontal velocity and final vertical velocity. Since these two components are perpendicular to each other, the magnitude of the final velocity \(v\) can be found using the Pythagorean theorem:
\(v = \sqrt{v_x^2 + v_y^2}\)
Substitute the values we found for \(v_x\) and \(v_y^2\):
\(v = \sqrt{u^2 + 2gh}\)
This formula gives the magnitude of the velocity of the dropped packet just before it hits the ground.
Thus, the velocity of the dropped packet on reaching the ground is \(\sqrt{u^2+2gh}\).
Important Questions from Motion in Two and Three Dimensions
A running truck at 54 km/hr is bought to rest in 10s. The distance travelled by the truck in 10th second is:
A cage in a mine-shaft descends with 2 ft/s2 units of acceleration. After it has been in motion for 10 seconds a particle is dropped on it from the top of the shaft with 32 ft/s2. What time elapses before the particle hits the cage?
An object is launched horizontally from a height of 45 m above the ground with a speed of 10 m/s. The time at which the object will hit the ground is : (Take g = 10 m/s2)
A car starts from Bengaluru, goes 50 km in a straight line towards south, immediately turns around and returns to Bengaluru. The time taken for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip
- Motion along a curved path and confined to one plane is known as ________.
