Motion Under Gravity, Free Fall Equations, Formulas, MCQs, Worksheet, Test Paper

Learn the complete concept of vertical motion, including the derivation of kinematic equations for bodies projected upwards, in free fall, and thrown from heights. Detailed explanations, formulas, solved examples, and exam-oriented questions for JEE, NEET, and CBSE Class 11.

Body Projected Vertically Upwards

Concept Explanation:

When a body is projected vertically upwards with an initial velocity u, it moves against gravity and reaches a maximum height H before falling back down.

• Acceleration due to gravity g acts downward.
• At the highest point, velocity becomes zero v = 0.

Using the equations of motion :

1. First Equation of Motion: $v = u – gt$. This equation is derived from the definition of acceleration: $$a = \frac{\text{change in velocity}}{\text{time}} = \frac{v – u}{t}$$ Since acceleration due to gravity is downward and motion is upward, $$a = -g$$ $$-g = \frac{v – u}{t}$$ Rearranging: $$v = u – gt$$
2. Second Equation of Motion: $h = ut – \frac{1}{2} gt^2$. Starting from the definition of displacement: $$s = ut + \frac{1}{2} at^2$$ Substituting a = -g: $$h = ut – \frac{1}{2} gt^2$$
3. Third Equation of Motion: $v^2 = u^2 – 2gh$. Using the first equation of motion: $$v = u – gt$$ Squaring both sides: $$v^2 = u^2 – 2gu t + g^2 t^2$$ From the second equation: $$t^2 = \frac{2h}{g}$$ Substituting this value and simplifying gives: $$v^2 = u^2 – 2gh$$

Important Results:

• Maximum height : $$H = \frac{u^2}{2g}$$ (derived by setting v = 0 in third equation)
• Time to reach maximum height : $$t = \frac{u}{g}$$ (derived from first equation with v = 0)
• Total time of flight : $$T = 2t = \frac{2u}{g}$$
• Velocity when returning to the ground : $$v = -u$$ (same magnitude but opposite direction)

Graphical Representation:

1. Displacement-time graph (parabolic curve)
2. Velocity-time graph (linear decrease to zero, then linear increase in the opposite direction)
3. Acceleration-time graph (constant negative -g)

Observations:

• Time of ascent = Time of descent $$t_a = t_d = \frac{u}{g}$$
• Speed at which the body is projected = Speed when it returns to the ground

Body Dropped from a Height

Concept Explanation:

When a body is dropped from a certain height H (initial velocity u = 0 at highest point of its motion), it accelerates downward under gravity a = g.

Using the equations of motion :

1. Velocity after time t : $$v = gt$$ Derived from the first equation of motion by setting u = 0.
2. Distance traveled in time t : $$H = \frac{1}{2} gt^2$$ Derived from the second equation of motion by setting u = 0.
3. Velocity before hitting the ground : $$v^2 = 2gH$$ Derived from the third equation by setting u = 0.
   • Hence, velocity of body just before hitting the ground is : $$v = \sqrt{2gH}$$
4. Time to reach the ground : Distance traveled by body in time t is $H = \frac{1}{2} gt^2$. Hence time taken by body to reach the ground is $$t = \sqrt{\frac{2H}{g}}$$

Important Results:

• Velocity before hitting the ground: $v = \sqrt{2gH}$
• Time to reach the ground: $t = \sqrt{\frac{2H}{g}}$

Graphical Representation:

1. Distance-time graph (parabolic curve)
2. Velocity-time graph (linear increase)
3. Acceleration-time graph (constant +g)

Observations:

• Distance covered in $n^{th}$ second: $h_n = \frac{1}{2} g(2n – 1)$
• Ratio of distances covered in successive seconds: 1 : 3 : 5 : 7
• Ratio of distances covered in time $t, 2t, 3t$ : $1^2:2^2:3^2$

Body Thrown Vertically from a Height

Concept Explanation:

There are two cases:

1. Upward throw from a height: The body first moves up (velocity opposite to gravity), reaches a highest point, then undergoes free fall.
2. Downward throw from a height: The body is thrown downward with initial velocity uu, meaning velocity and acceleration are in the same direction. It reaches the ground faster than free fall.

Equations Used:

• Assume appropriate signs for motion direction.
• Use standard kinematic equations based on given conditions.

Quick Revision Points:

• Free fall motion has u = 0, a = g.
• The time taken to go up equals the time taken to come down.
• Velocity at any point is the same in magnitude during ascent and descent.
• The distance covered in $n^{th}$ second follows the odd number rule.

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