Projectile Motion - Free MCQ Practice Test with solutions, NEET Physics

A projectile is any object thrown into space upon which the only acting force is gravity. A bullet fired from a gun, a kicked football, and a javelin thrown by an athlete are examples of projectiles once they are in motion and only gravity acts on them. Taking off of an aircraft is not a projectile because it involves continuous external force from the engines.

Ship 2 gets hit first. The time of flight only depends on the y-component of motion, not the x-component.  The higher you throw something up in the air, the more time it spends in the air.  It can also be shown from equation-

y=v_0y t – ½ g t²=0  

    v_0y for projectile μ1 is greater than for μ2.

When an object is still held it is the part of a system in which the train is moving, once it is left it still has the same velocity as the train for a person from ground. Thus as the velocity is horizontal and acceleration is vertical, it would follow a parabolic trajectory.

At the highest point in its trajectory, the acceleration (g) is acting in downward direction, whereas velocity of the projectile is only in the X (horizontal) direction.

We know that the horizontal range for any projectile motion let say R = 2u².sin 2a /g
Where u is initial speed, and a is the angle at which the particle is thrown, which is responsible for direction. So in between the possible range of a that is 0 - 90, there are maximum two equal values of sin 2a, thus the maximum number of directions for the same or equal range are 2.

At the highest point, its vertical velocity will become zero. Hence the only left velocity is horizontal which remains unchanged and is equal to 10 cos 60 = 5m/s

As there is no horizontal acceleration is present in oblique projection horizontal component of velocity is constant

At a point of maximum height, the derivative of displacement i.e. velocity is zero but as the gravitational acceleration is equal at all near points to the surface of earth, acceleration at maximum height is still equal to g.

Sine of an angle has maximum value 1 when the angle is 90 degree Rmaxis obtained when 2ø = 90 degreeorø= 45 degree

Let the initial speed be u. The vertical component of initial velocity is u‑sinθ, and the acceleration is g downward.

Time of ascent (time to reach maximum height) is found from v = u‑sinθ - g t with final vertical velocity v = 0, giving t₁ = (u‑sinθ)/g.

The maximum height is h = (u² sin²θ)/(2g). Time to fall from this height to the launch level is t₂ = √(2h/g) = (u‑sinθ)/g, so t₂ = t₁.

Therefore, when the projectile returns to the same vertical level from which it was projected, the times of rise and fall are equal. If the launch and landing heights are different, the rise and fall times are not generally equal.