PhysicsChapter 3: Translatory Motion
Free-Fall Motion
19. Definition of Free Fall
20. Acceleration Due to Gravity
Near Earth’s surface, all freely falling objects have approximately the same downward acceleration: g ≈ 9.8 m s⁻² For simplified numerical questions, the value may be given as: g = 10 m s⁻² The direction of g is always vertically downward. An object’s mass does not affect its free-fall acceleration when air resistance is ignored.
21. Free-Fall Assumptions
Standard free-fall equations normally assume that:
- Air resistance is negligible.
- Gravity is the only force acting on the object.
- The value of g remains constant.
- Motion occurs close to Earth’s surface.
- The object moves vertically.
22. Free-Fall Sign Convention
A positive direction must be selected before solving the problem.
When Upward Is Positive
a = −g For an object projected upward: v = u − gt s = ut − ½gt² v² = u² − 2gs
When Downward Is Positive
a = +g For an object dropped from rest: u = 0 v = gt h = ½gt² v² = 2gh
Free-Fall Motion
23. Important Free-Fall Cases
Object Dropped From Rest
u = 0 Taking downward as positive: v = gt h = ½gt² v² = 2gh
Object Projected Vertically Upward
Taking upward as positive: a = −g v = u − gt s = ut − ½gt² v² = u² − 2gs
At the Highest Point
v = 0 However: a = −g The velocity is momentarily zero, but acceleration is not zero.
24. Time to Reach Maximum Height
At maximum height: v = 0 Using: v = u − gt Substituting v = 0: 0 = u − gt gt = u tup = ug If the object returns to the same level: ttotal = 2u / g
25. Maximum Height
At maximum height: v = 0 Using: v² = u² − 2gh Substituting v = 0: 0 = u² − 2gh 2gh = u² h_max = u²2g
26. Solved Example 4: Object Dropped From Rest
Solved Example 4: Object Dropped From Rest
An object is dropped from a height of 45 m. Calculate its time of fall and final velocity. Take g = 10 m s⁻².
- u = 0
- h = 45 m
- g = 10 m s⁻²
- Downward is taken as positive.
Time of Fall: h = ½gt²
45 = ½(10)t²
45 = 5t²
t² = 9
t = 3 s
Final Velocity: v = gt
v = (10)(3)
v = 30 m s⁻¹
Final answer: t = 3 s; v = 30 m s⁻¹ downward
27. Solved Example 5: Object Projected Upward
Solved Example 5: Object Projected Upward
A ball is projected vertically upward with an initial velocity of 20 m s⁻¹. Calculate its time to reach maximum height, maximum height, and total time of flight. Take g = 10 m s⁻².
- u = 20 m s⁻¹
- g = 10 m s⁻²
- At maximum height: v = 0
Time to Reach Maximum Height: tup = ug = 2010 = 2 s
Maximum Height: hmax = u²2g = (20)² / (2 × 10) = 40020 = 20 m
Total Time: ttotal = 2 tup = 2(2) = 4 s
Final answer: tup = 2 s; hmax = 20 m; ttotal = 4 s
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