Understanding the Motion of a Ptopic Thrown from a Tower of Height hWhen a ptopic is thrown from the top of a tower, its motion follows a specific path determined by gravity, initial velocity, and the height from which it is launched. This scenario is a classic example of projectile motion in physics. Understanding how the ptopic moves helps us analyze key physical quantities such as time of flight, horizontal range, and final velocity.
What Is Projectile Motion?
Projectile motion is the curved path that an object follows when it is thrown or projected near the surface of the Earth and is influenced only by gravity (neglecting air resistance). It involves two types of motion at once
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Horizontal motion with constant velocity.
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Vertical motion with uniform acceleration due to gravity.
When a ptopic is projected from a height, like from a tower, it doesn’t fall straight down unless dropped vertically. If thrown with some horizontal or angular velocity, it travels along a curved trajectory.
Types of Projections from a Tower
There are three main ways a ptopic can be projected from the top of a tower
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Vertically Downward
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Horizontally
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At an Angle to the Horizontal
Each case affects the time taken, distance covered, and final speed differently.
Case 1 Ptopic Thrown Vertically Downward
If the ptopic is thrown straight down from a tower of height h with an initial speed u, only vertical motion is involved.
Equations Used
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Displacement s = ut + frac{1}{2}gt^2
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Solving for time h = ut + frac{1}{2}gt^2
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Final velocity v = u + gt
This case results in the shortest time of fall compared to other cases with the same height, assuming u > 0.
Case 2 Ptopic Thrown Horizontally from a Tower
This is one of the most common physics problems. The ptopic has an initial horizontal velocity u but no vertical velocity at the start.
Key Characteristics
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Horizontal motion Constant velocity, x = ut
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Vertical motion Accelerated due to gravity, y = frac{1}{2}gt^2
Time of Flight
- The time taken to hit the ground depends only on the height h t = sqrt{frac{2h}{g}}
Horizontal Range
- The total distance traveled horizontally is R = u cdot t = u cdot sqrt{frac{2h}{g}}
Final Speed
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The vertical velocity just before hitting the ground is v_y = gt
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The total final speed is v = sqrt{u^2 + (gt)^2}
Case 3 Ptopic Thrown at an Angle from a Tower
If the ptopic is thrown at an angle θ with initial speed u, it has both horizontal and vertical components
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u_x = u costheta
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u_y = u sintheta
Now, both components affect the motion.
Time of Flight
This requires solving the vertical motion equation with initial vertical velocity
h + u sintheta cdot t – frac{1}{2}gt^2 = 0
Solving this quadratic equation gives the total time the ptopic stays in the air.
Horizontal Range
R = u costheta cdot t
The range depends on the horizontal velocity and the total time of flight.
Maximum Height Above the Tower
This is given by
H = frac{(u sintheta)^2}{2g}
The maximum height reached from the ground is then
h + H
Real-Life Applications
Engineering and Architecture
Understanding how objects fall from height helps in structural design and safety engineering, such as placing safety nets or calculating fall trajectories.
Sports
Many sports like basketball or javelin throw depend on correctly judging angles and initial speeds, which are essentially projectile motion problems.
Space and Ballistics
Even in designing rocket launches or determining missile paths, the concepts of motion from a height are crucial.
Factors Affecting the Ptopic’s Path
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Initial Velocity (u) Determines how far the ptopic will travel horizontally.
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Angle of Projection (θ) Affects the height, time of flight, and range.
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Height of Tower (h) A taller tower increases the time the ptopic remains in the air.
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Acceleration due to Gravity (g) On Earth, g is approximately 9.8 m/s², but it would vary on other planets.
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Air Resistance While typically neglected in basic problems, it can alter the actual path in real-world scenarios.
Sample Problem
Problem A ptopic is thrown horizontally from a tower 80 meters high with an initial speed of 10 m/s. Find the time of flight, range, and final speed.
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Time of flight t = sqrt{2h/g} = sqrt{160/9.8} approx 4.04 text{ s}
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Horizontal range R = u cdot t = 10 cdot 4.04 = 40.4 text{ m}
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Final vertical speed v_y = g cdot t = 9.8 cdot 4.04 approx 39.6 text{ m/s}
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Final speed v = sqrt{10^2 + 39.6^2} approx 40.8 text{ m/s}
Throwing a ptopic from a tower of height h is a classic example of projectile motion, showcasing how gravity and initial velocity interact to define the ptopic’s path. By analyzing the scenario with basic physics principles, one can determine important values such as time of flight, horizontal range, and final speed.
These insights are valuable not just in solving academic problems but also in understanding the physical behavior of objects in motion across various fields including engineering, sports, and space science.