Motion from a tower: dropped, thrown up or down
The tower problem: initial velocity may be zero, up, or down — treat the whole motion as one parabola in time.
In this lesson
- The tower problem: initial velocity may be zero, up, or down — treat the whole motion as one parabola in time.
A ball thrown UP from a cliff still falls past the cliff edge — sign convention decides everything.
Take the tower top as origin, down as positive. Dropped: u = 0. Thrown down: u > 0. Thrown up: u < 0 (up is negative). Then s = ut + ½gt² with s = +h (the tower height) at landing.
Solve the quadratic for t; the positive root is the answer. The negative root is the unphysical 'before the throw' time.
Tower-top origin, down positive
Worked example
The acceleration of a body starting from rest is a = 2 m/s² for 4 s, then zero for 2 s. Find its velocity at t = 6 s.
- Area under the a–t graph = change in velocity.
- Area = 2 m/s² × 4 s = 8 m/s (the last 2 s add nothing).
Answer: 8
Transcript (0 min)
WEBVTT 1 00:00:00.000 --> 00:00:05.000 Motion from a tower: dropped, thrown up or down — FemtoLearn. 2 00:00:05.000 --> 00:00:15.000 The tower problem: initial velocity may be zero, up, or down — treat the whole motion as one parabola in time.
Practice
Free · 4 questions with full solutions- Q1 · Numerical · difficulty 2/5
A stone is dropped from rest from a height of 10 m. Taking g = 10 m/s², find the time it takes to reach the ground (neglect air resistance).
- Q2 · Numerical · difficulty 2/5
The acceleration of a body starting from rest is a = 2 m/s² for 4 s, then zero for 2 s. Find its velocity at t = 6 s.
- Q3 · MCQ · difficulty 2/5
A particle moves with constant velocity. Which a–t graph describes it?
- A horizontal line at a = 0
- A horizontal line at a = 2 m/s²
- A straight line through the origin
- A parabola
- Q4 · Numerical · difficulty 2/5
A body moves with v = 10 m/s for 3 s, then v = 5 m/s for 2 s in the same direction. Find the total distance.