Question

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two objects are fired into the air. the drawing shows that they reach the same height, but the ranges of their projectile motions are different. which one is in the air for the greatest amount of time?
projectile 1, because it has the smaller range and, hence, the horizontal component of its velocity is smaller than that of projectile 2.
projectile 2, because it has the greater range.
both projectiles are in the air for the same amount of time.
projectile 1, because it has the smaller initial speed and, therefore, travels more slowly than projectile 2.

Explanation:

Step1: Analyze vertical motion

The time - of - flight of a projectile is determined by its vertical motion. The vertical displacement \(y - y_0\) of a projectile is given by \(y - y_0=v_{0y}t-\frac{1}{2}gt^{2}\), and at the maximum height \(v_y = v_{0y}-gt = 0\), so the time to reach the maximum height \(t_{up}=\frac{v_{0y}}{g}\), and the total time of flight \(T = \frac{2v_{0y}}{g}\).

Step2: Consider the given condition

Since the two projectiles reach the same height, the vertical component of their initial velocities \(v_{0y}\) is the same (using \(v_y^{2}-v_{0y}^{2}=- 2g(y - y_0)\), when \(v_y = 0\) at the maximum - height \(h\), \(v_{0y}=\sqrt{2gh}\)).

Step3: Determine the time of flight

Using the formula \(T=\frac{2v_{0y}}{g}\), because \(v_{0y}\) is the same for both projectiles, the time of flight \(T\) is the same for both projectiles.

Answer:

Both projectiles are in the air for the same amount of time.