Horizontal component of velocity would cause the bomb to miss the target, as it would cause the bomb to deviate horizontally from its intended impact point. The vertical component of velocity and pull of gravity would both cause the bomb to fall towards the target,

At the highest point of the basketball's trajectory, its vertical velocity becomes zero. However, gravity continues to act on the ball, causing it to decelerate and eventually start moving downward.

The acceleration at the highest point is equal to the acceleration due to gravity
(g), but it acts in the downward direction.

The distance of a broad jump is determined by the horizontal velocity and take off angle of the jumper at the time of launch

At the highest point,

       The horizontal component of velocity = v cos 

       The vertical component of velocity = 0

Newton-second (N·s) is the unit of impulse, which is a measure of the change of momentum of an object. Momentum is a measure of the motion of an object, and it is equal to the product of an object's mass and velocity.

Range will be maximum when:

• Θ=45
• 2Θ=90
  or
• sin2Θ=1

Since no external force is acting on the bullet + Tree trunk system when a bullet gets embedded in the Tree trunk, thus total momentum of the system is conserved.

But total kinetic energy of the system is not conserved as it is a perfectly inelastic collision and some kinetic energy is lost as heat.

To improve the jumping record, the longest jumper should jump at an angle close to 45 degrees for optimal performance.

2gh = Vf²Vi²
h= V²/2g
h= (15)²/2g
h= 11m

Two angles are called complementary when their measures add to 90 degrees. Two angles are called supplementary when their measures add up to 180 degrees.

Projectile possess minimum kinetic energy at the highest point of the trajectory i.e. at a horizontal distance R/2

In projectile thrown at angle θ Range R and maximum height H are given as :
R tanθ=4H now R=H tanθ=4H/R tanθ=4H/H tanθ=4 θ=tan¯¹(4) θ=76°

Use the Following Formula to Solve this Question
t=√(2h/g)
t=√(2(490)/10)=10 sec

As the magnitude and direction of velocity is constant hence net force is equal to zero.

When the stones were thrown from the top of a tall building due to the presence of gravity it follows the projectile motion. The projectile motion follows the trajectory of a parabolic path throughout the journey.

Two angles are called complementary when their measures add to 90 degrees. Two angles are called supplementary when their measures add up to 180 degrees.

Use the following formula to solve such a MCQS

T=2vsinθ/g
T=2(98)(0.5)/9.8=10 Sec

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Due to constant velocity along horizontal and vertical downward force of gravity ball will hit the ground following parabolic path.

Acceleration of the ball during any stage is constant in g and direction ,it is g=9.8m/s^2
and its downwards.

A projectile such as a ball thrown undergoing a parabolic path is made of two kinds of motion. They are the horizontal motion and the vertical motion. The horizontal motion is a uniform motion. It has no acceleration while the vertical motion is an accelerated motion. The direction of the acceleration is always towards the ground.

Individually it is always 9.8m/s^2 and acceleration of one relative to other is zero.

Extra Information: 

When train have some initial velocity and acceleration, and if ball if thrown in air it will carry only velocity of train and will behave as a freely falling body.

Now initial horizontal velocity of both the bodies is same and train have some acceleration in horizontal velocity but ball dont have any acceleration in horizontal.

So displacement of train is more in horizontal as compare to ball, so ball will land behind the thrower.

Change in velocity per unit time, is correct because both objects will experience the same acceleration due to gravity regardless of their mass. This acceleration is known as the acceleration due to gravity, or g, and it is equal to 9.8 m/s^2 on the surface of the Earth. As a result, the change in velocity per unit time, or acceleration, will be the same for both objects.

Any body in free fall is under acceleration due to gravity (g) and thus accelerates at $9.8m/s^2$. When a body starts from rest, after one second it will gain an acceleration of $9.8m/s^2$ and thus will be at a velocity of $9.8m/s$ in the downward direction

The vertical component of acceleration is g and because acceleration is constant (9.81ms^-2) throughout so it would be the same at anywhere on the parabola

AS WE Know at maximum height the vertical component is Zero but the horizenal component still there which is Vx=VCos@

1..When the projectile is launched from a height horizontally, the vertical motion is one of constant downward acceleration.
2..The horizontal component remains constant throughout the motion.
3..Therefore, the velocity of a projectile equals its initial velocity when added to a constantly increasing downward velocity.