When a body is dropped from a tower, its velocity increases continuously due to the gravitational force pulling it downward.

This acceleration causes the body to gain speed as it falls. At the same time, the body's potential energy decreases because it is converting into kinetic energy—the energy of motion—as the body descends.

Explanation:
Initial velocity (a) = 20 km/h; Final velocity (v) = 60 km/h and time (t) = 4 hours. velocity (v) =  60 = u + at = 20 + (a x 4)

Therefore distance travelled in 4 hours is s

When a particle is moving with uniform velocity then its speed also zero and acceleration become zero.

The position vector of the particle changes as time progresses, even if the particle is moving with uniform velocity, because its location in space is continuously changing.

During the flight of an oblique projectile:

• The horizontal velocity component (Vx) remains constant.
• The vertical velocity component (Vy) first decreases, reaches a minimum, and then increases.

Hint:  
and R both are proportional to u² Hence, percentage increase in horizontal range would also be 10%

Hint:
Hence Velocity is Horizontal and Acceleration is Vertical

Hint: An External Force by gravity is Present Throughout the Motion so Momentum cannot Be Conserved

Only the horizontal component of velocity and acceleration remain constant for a projectile, as they are not affected by the vertical forces of gravity.

Explanation of Incorrect Options:

Vertical component of velocity and kinetic energy: Both change continuously due to gravity.

Potential energy and kinetic energy: They constantly convert from one form to another throughout the projectile's motion.

Potential energy and acceleration: Acceleration is always due to gravity, which affects both potential and kinetic energy, making them change.

Acceleration perpendicular to motion: Since the acceleration is always perpendicular to the object's motion, it doesn't cause any change in the magnitude of the velocity (speed). Imagine a car turning a corner – its speed stays the same, but its direction changes.

Change in direction: However, the continuous perpendicular acceleration constantly alters the direction of the velocity. Think of a ball rolling on a circular track – its speed remains constant, but it keeps changing direction to follow the curve.

Since the acceleration due to gravity is vertically downward, at the highest point Q, where the direction of motion is to the right (horizontal velocity remains constant since air resistance, which opposes motion, is negligible [also, it is not affected by the vertically down acceleration], so the horizontal component of acceleration is zero).

At the highest point Q, there is no vertical upward component since the projectile has reached its maximum vertical height – it cannot go any higher.

The horizontal component of the projectile’s velocity is not zero since the motion continues towards the right (note that zero acceleration does not mean a speed of zero – it just means that the speed is not changing). [B is incorrect]

Both the kinetic energy and momentum of the projectile depends on its speed. Since the speed is not zero, the kinetic energy and momentum cannot be zero. [C and D are incorrect]

The slope of a velocity-time graph directly represents the object's acceleration at any given time.

Area under velocity-time graph: Gives total displacement, not instantaneous acceleration.

Slope of displacement-time graph: Gives velocity, not acceleration.

The balloon is moving upwards while the sandbag is falling downwards. We have motion in both the upward and downward directions.

The speed and time graph of free falling body is straight line so option C. & D. are not correct.
Now,
The distance and time relation of a free falling body is given by S=1/2 at^2
Where S is distance, a is acceleration and  t is time
Now when the air resistance is negligible the graph is like

So option (A) is also not correct this the Correct option is option B. Where rate of change of Distance is decreasing with time.

A force can cause a stationary object to start moving, a moving object to decrease speed, and a moving object to change its direction of motion.

To find the magnitude of the displacement of the car, we must represent each velocity of the car as a vector. We will then add the vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

The cannon ball has a two-dimensional motion, specifically projectile motion. Any two-dimensional motion consists of two one-dimensional motions happening at the same time. This is why we break vectors into components, so we can analyze each one-dimensional motion separately. The motion in the y-axis and the x-axis are connected by the time variable, since they happen at the same time. For projectile motion, we have that on the y-axis (the vertical axis) a constant acceleration (gravity) is acting on the cannon ball. For the y-axis motion with constant acceleration or free fall (which is motion with constant acceleration where the acceleration is gravity) are correct choices. On the x-axis (the horizontal axis), there is no acceleration acting on the cannon ball horizontally. Therefore, on the x-axis the cannon ball experiences motion with constant velocity. Then the only answer choice that describes the motion of the cannon ball in each axis correctly is: y-axis: motion with constant acceleration and x-axis: motion with constant velocity.

To find the average velocity, the displacement needs to be measured. The displacement the difference between The final position and the initial position.

75m−50m=25m

Using the equation for velocity, velocity=displacement/time, we can find the average velocity.

v=25/5=5ms

We are given the time that the ball is in the air and the acceleration. Since the ball is dropped, we can also tell that the initial velocity is zero (starts from rest). Using these given values, we can calculate the final velocity using the appropriate motion equation.

The final velocity is negative because the ball is traveling downward; however, the question asks for speed, allowing us to disregard the vector indicator. Speed cannot be negative.

We are given the time that the ball is in the air and the acceleration. Since the ball is dropped, we can also tell that the initial velocity is zero (starts from rest). Using these given values, we can calculate the final velocity using the appropriate motion equation.

The resulting displacement is negative because the ball is traveling downward; however, the height will be the absolute value of the displacement, giving out final answer.

In the case the motion is uniform velocity motion then the graph between velocity and time is a straight line which indicates that the slope is zero$.$ Normally slope of velocity$-$time graph is acceleration but when it approaches a particular point then it becomes instantaneous acceleration$.$

If the gradient of a graph is negative, then the acceleration is "negative." The negative gradient indicates a decrease in velocity over time, corresponding to deceleration or negative acceleration

Explanation

Gravitational Force Act downward but also Fg=mg