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Physics Chapter 3 : Topic 2
Projectile Motion
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Projectile motion is a type of motion in which a body:
Reasoning: A projectile is launched with an initial velocity and then moves under gravitational acceleration.
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In projectile motion without air resistance, the horizontal acceleration is:
Reasoning: No horizontal force acts on the projectile, so horizontal acceleration is zero.
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The only force acting on an ideal projectile is:
Reasoning: In the absence of air resistance, only gravitational force acts.
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The path followed by a projectile is:
Reasoning: Combination of horizontal uniform motion and vertical accelerated motion gives a parabolic path.
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The acceleration of a projectile during its entire flight is:
Reasoning: Gravitational acceleration remains constant downward.
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A projectile launched with velocity ‘$\mathbf{u}$’ at angle $\mathbf{\theta}$ has horizontal velocity component:
Reasoning: Horizontal component is the adjacent component of velocity.
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The initial vertical component of projectile velocity is:
Reasoning: Vertical component is obtained using sine.
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If a projectile is launched horizontally, its initial vertical velocity is:
Reasoning: A horizontal launch has no initial upward or downward velocity.
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At an angle of $\mathbf{45}^{\mathbf{o}}$, the horizontal and vertical components of velocity are:
Reasoning: $\mathbf{u}_{\mathbf{x}}\mathbf{= u\ cos\ }\mathbf{45}^{\mathbf{o}}$ ; $\mathbf{u}_{\mathbf{y}}\mathbf{= u\ sin\ }\mathbf{45}^{\mathbf{o}}$ $\mathbf{\Longrightarrow}$ Both are equal.
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Increasing the launch angle while keeping velocity constant increases the:
Reasoning: Increasing angle increases the vertical component and decreases horizontal component.
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The horizontal displacement of a projectile after time ‘$\mathbf{t}$’ is:
Reasoning: Horizontal motion is uniform.
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The horizontal velocity of an ideal projectile:
Reasoning: No horizontal acceleration acts.
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If the horizontal component of velocity is doubled, the horizontal range becomes:
Reasoning: Range depends directly on horizontal velocity.
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The vertical motion of a projectile is similar to:
Reasoning: Vertical motion is controlled by gravitational acceleration.
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At maximum height of a projectile:
Reasoning: The projectile momentarily stops moving upward.
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The acceleration of a projectile at the highest point is:
Reasoning: Gravity acts throughout the motion.
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The maximum height of a projectile depends on:
Reasoning: Maximum height is determined by upward velocity.
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The formula for maximum height is:
Reasoning: Vertical velocity determines height.
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The total time of flight of a projectile launched from ground is:
Reasoning: $\mathbf{T =}\frac{\mathbf{2}\mathbf{u}_{\mathbf{y}}}{\mathbf{g}}$ $\mathbf{\Longrightarrow}$ $\mathbf{T =}\frac{\mathbf{2u\ sin\ \theta}}{\mathbf{g}}$
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Increasing launch angle (below $\mathbf{90}^{\mathbf{o}}$) generally increases:
Reasoning: Larger vertical component keeps projectile in air longer.
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The time of flight depends on:
Reasoning: Vertical motion determines how long the projectile remains airborne.
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The horizontal range of a projectile is:
Reasoning: Projectile range formula.
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The range of a projectile is maximum when the launch angle is:
Reasoning: $\mathbf{sin\ 2\theta}$ becomes maximum when $\mathbf{2\theta =}\mathbf{90}^{\mathbf{o}}$.
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A projectile launched at $\mathbf{30}^{\mathbf{o}}$ and $\mathbf{60}^{\mathbf{o}}$ with the same velocity has:
Reasoning: Complementary angles produce the same range.
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The angles producing the same range are:
Reasoning: $\mathbf{\theta}_{\mathbf{1}}\mathbf{+}\mathbf{\theta}_{\mathbf{2}}\mathbf{=}\mathbf{90}^{\mathbf{o}}$
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A projectile launched at $\mathbf{90}^{\mathbf{o}}$ has:
Reasoning: No horizontal component exists.
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A projectile is launched with speed $\mathbf{20\ m/s}$ at $\mathbf{30}^{\mathbf{o}}$. Its horizontal component is:
Reasoning: $\mathbf{u}_{\mathbf{x}}\mathbf{= 20\ cos\ }\mathbf{30}^{\mathbf{o}}\mathbf{= 17.3\ m/s}$
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A projectile is launched with speed $\mathbf{20\ m/s}$ at $\mathbf{30}^{\mathbf{o}}$. Its vertical component is:
Reasoning: $\mathbf{u}_{\mathbf{y}}\mathbf{= 20\ sin\ }\mathbf{30}^{\mathbf{o}}\mathbf{= 10\ m/s}$
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A projectile has vertical velocity $\mathbf{2}\mathbf{0\ m/s}$. Taking $\mathbf{g = 10\ m/}\mathbf{s}^{\mathbf{2}}$, maximum height is:
Reasoning: $\mathbf{H =}\frac{\mathbf{u}_{\mathbf{y}}^{\mathbf{\ 2}}}{\mathbf{2g}}$ $\mathbf{\Longrightarrow}$ $\mathbf{H =}$ $\frac{\mathbf{400}}{\mathbf{20}}\mathbf{= 20\ m}$
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A projectile remains in air for $\mathbf{4}$ seconds. If $\mathbf{g = 10\ m/}\mathbf{s}^{\mathbf{2}}$, its initial vertical velocity is:
Reasoning: $\mathbf{T =}\frac{\mathbf{2}\mathbf{u}_{\mathbf{y}}}{\mathbf{g}}$ $\mathbf{\Longrightarrow}$ $\mathbf{4 =}$ $\frac{\mathbf{2}\mathbf{u}_{\mathbf{y}}}{\mathbf{10}}$ $\mathbf{\Longrightarrow}$ $\mathbf{u}_{\mathbf{y}}\mathbf{= 20\ m/s}$
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At any point during projectile motion, horizontal and vertical motions:
Reasoning: The two components occur independently.
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If air resistance is ignored, the mass of projectile affects:
Reasoning: All ideal projectile properties are independent of mass.
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Two projectiles with the same velocity but different masses have:
Reasoning: Mass does not affect projectile motion without air resistance.
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The velocity of a projectile at the highest point is:
Reasoning: Only vertical component becomes zero.
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If the initial speed of a projectile is increased, its maximum height:
Reasoning: Height depends on square of initial velocity.
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The range of a projectile is zero when:
Reasoning: No horizontal velocity exists at $\mathbf{90}^{\mathbf{o}}$.
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A projectile launched at $\mathbf{45}^{\mathbf{o}}$ has:
Reasoning: $\mathbf{45}^{\mathbf{o}}$ gives maximum range.
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When a projectile returns to the same height from which it was launched:
Reasoning: Without air resistance, speed remains the same at equal heights.
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The velocity direction of a projectile at maximum height is:
Reasoning: Only horizontal velocity remains.
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A projectile motion problem should usually be solved by:
Reasoning: Projectile motion is analyzed by resolving motion into independent components.
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