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Kinematics & dynamics · Lesson 2 of 3

Motion under gravity and motion from a formula

Bodies thrown up, dropped or thrown down, with g as the acceleration; then motion given by a formula, using differentiation and integration.

18 minYou should already know: Vectors Calculus (JAMB bridge)
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Motion under gravity

A body moving freely up or down has a constant acceleration gg (take g=10 m s−2{g = 10\text{ m s}^{-2}} unless told otherwise), always downwards. So the equations of motion apply, with a=−ga = -g if you take up as positive.

uv = 0 at the topgdown againground
Thrown straight upAt the top v = 0; greatest height = u² ÷ 2g
  • At the highest point the velocity is 00 for an instant.
  • The time up equals the time back down to the same level.
  • A body passes a given height twice: once going up, once coming down.

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q15

A particle is projected vertically upwards from the ground with speed 30 m s−130\text{ m s}^{-1}. Calculate the: (i) maximum height reached by the particle; (ii) time taken by the particle to return to the ground; (iii) time(s) taken for the particle to attain a height of 40 m40\text{ m} above the ground. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

  1. Greatest height

    • v2=u2−2gs{v^2 = u^2 - 2gs}: 0=900−20s{0 = 900 - 20s}.
    • So s=45{s = 45} m.

    Think first. At the top v = 0. Which equation has no t?

  2. Back to the ground

    • 0=30−10t{0 = 30 - 10t}, so it takes 3{3} s to go up.
    • It takes as long to come down, so it lands after 6{6} s.

    Think first. How long does it take to reach the top?

  3. At 40 m

    • 40=30t−5t2{40 = 30t - 5t^2}, so t2−6t+8=0{t^2 - 6t + 8 = 0}.
    • (t−2)(t−4)=0{(t - 2)(t - 4) = 0}: at t=2{t = 2} s (going up) and t=4{t = 4} s (coming down).

    Think first. Put s = 40 into s = ut − ½gt².

When a body starts above the ground, take up as positive and treat the ground as a negative displacement:

Thrown up from a heightSet the speed and height, then slide the time
0123452040t (s)h (m)
30 mheight5 m/svelocity (up +)4 slands at
h = 20 + 15t − 5t² and v = 15 − 10t. Highest at t = 15 ÷ 10 = 1.5 s, at 31.25 m. It lands when 20 + 15t − 5t² = 0: t = 4 s. Going up: v is positive.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q8

A ball is thrown vertically upwards with a velocity of 10 m s−110\ \text{m s}^{-1} from a point 7575 metres above the ground. How long will it take the ball to strike the ground? [Take g=10 m s−2][\text{Take } g = 10\ \text{m s}^{-2}]

Time in seconds

  1. Signs

    • u=+10{u = +10}, a=−10{a = -10}, and the ground is 75 m below, so s=−75{s = -75}.

    Think first. Up is positive. What is the displacement when it lands?

  2. Solve

    • s=ut+12at2{s = ut + \frac12at^2}: −75=10t−5t2{-75 = 10t - 5t^2}.
    • t2−2t−15=0{t^2 - 2t - 15 = 0}, so (t−5)(t+3)=0{(t - 5)(t + 3) = 0}.
    • Time cannot be negative, so t=5{t = 5} s.

More: motion under gravity

Motion from a formula

When the displacement, velocity or acceleration is given as a formula in tt, use calculus (see calculus). Differentiate to go from ss to vv to aa; integrate to go back, using the starting values to find the constant.

svadifferentiate (d/dt)integrate (∫ dt)
Displacement, velocity, accelerationv = ds/dt, a = dv/dt; v = ∫a dt, s = ∫v dt

A body is momentarily at rest when v=0v = 0; for a body thrown up, that is at its greatest height.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q8

A body is thrown vertically upwards. Its height hh metres at time tt seconds is given by h=(12t−25t2) mh = \left(12t - \frac25t^2\right)\text{ m}. Find:

the time at which it is momentarily at rest;

the maximum height reached by the body.

  1. At rest: v = 0

    • v=dhdt=12−45t{v = \frac{dh}{dt} = 12 - \frac45t}.
    • 12−45t=0{12 - \frac45t = 0}, so t=15{t = 15} s.

    Think first. Differentiate h to get v.

  2. The greatest height

    • h=12(15)−25(15)2=180−90=90{h = 12(15) - \frac25(15)^2 = 180 - 90 = 90} m.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q8

The acceleration, aa, of a particle starting from rest and moving at any time tt seconds is given by a=(20t−3t2) m s−2a = (20t - 3t^2)\text{ m s}^{-2}. Find the:

time taken for the particle to come to rest again;

distance covered by the particle after 5 seconds.

  1. Velocity

    • v=∫(20t−3t2) dt=10t2−t3+c{v = \int (20t - 3t^2)\,dt = 10t^2 - t^3 + c}.
    • At t=0{t = 0}, v=0{v = 0}, so c=0{c = 0}.

    Think first. Integrate a. It starts from rest: what is the constant?

  2. At rest again

    • 10t2−t3=0{10t^2 - t^3 = 0}, so t2(10−t)=0{t^2(10 - t) = 0}.
    • Not the start: t=10{t = 10} s.

  3. Distance in 5 s

    • s=∫05(10t2−t3) dt=[10t33−t44]05{s = \int_0^5 (10t^2 - t^3)\,dt = \left[\frac{10t^3}{3} - \frac{t^4}{4}\right]_0^5}.
    • =12503−6254=312512≈260.42{= \frac{1250}{3} - \frac{625}{4} = \frac{3125}{12} \approx 260.42} m.

    Think first. v stays positive from 0 to 10, so integrate v from 0 to 5.

More: motion from a formula

Your turn

WAEC 2017 · Paper 2 · Q14 (a)

  1. (a)

    A particle is projected vertically upwards from a point OO with a velocity of 75 m s−175\text{ m s}^{-1}. Find the: (i) velocity of the particle at the end of 5 seconds; (ii) height attained when the velocity is 15 m s−115\text{ m s}^{-1}; (iii) times when the particle is 270 m270\text{ m} above OO. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. v=u−gtv = u - gt
    =75−10(5)= 75 - 10(5)
    =25 m s−1= 25\text{ m s}^{-1}.

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