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There are many real time situations that resemble horizontal projection. When an object is dropped from a plane flying parallel to the ground at certain height, then the object acquires horizontal velocity of the plane when the object is released. As the object is simply dropped, the velocity in vertical direction is zero. This horizontal velocity of the object, as acquired from the plane, is then modified by the force of gravity, whereby the object follows a parabolic trajectory before hitting the ground.
This situation is analogous to projection from ground except that we track motion from the highest point. Note that vertical velocity is zero and horizontal velocity is tangential to the path at the time of projection. This is exactly the same situation as when projectile is projected from the ground and reaches highest point. In the nutshell, the description of motion here is same as the description during descent when projected from the ground.
The interesting aspect of the object dropped from plane is that both plane and object are moving with same horizontal velocity. Hence, plane is always above the dropped object, provided plane maintains its velocity.
The case of projection from a higher level at certain angle (up or down) to the horizontal is different to the one in which projectile is projected horizontally. The projectile has a vertical component of initial velocity when thrown at an angle with horizontal. This introduces the difference between two cases. The projectile thrown up attains a maximum height above the projection level. On the return journey downward, it travels past its level of projection. The difference is visually shown in the two adjoining figures below.
The resulting trajectory in the first case has both upward and downward motions. On the other hand, the motion in upward direction is completely missing in the horizontal projection as the projectile keeps loosing altitude all the time.
We can easily analyze projectile motion following the technique of component motions in two mutually perpendicular directions (horizontal and vertical). Typically, we consider vertical component of motion to determine time of flight (T). The initial velocity in vertical direction is zero.
We consider point of projection as origin of coordinate system. Further, we choose x-axis in horizontal direction and y-axis in the vertically downward direction for the convenience of analysis. Then,
$$y=H={u}_{y}T+\frac{1}{2}g{T}^{2}$$
But ${u}_{y}=0$ ,
$$\Rightarrow H=\frac{1}{2}g{T}^{2}$$
$$\Rightarrow T=\sqrt{\left(\frac{2H}{g}\right)}$$
Note the striking similarity here with the free fall of a body under gravity from a height “H”. The time taken in free fall is same as the time of flight of projectile in this case. Now, the horizontal range of the projectile is given as :
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