Worked Examples-Calculation Part 2 - Motion in a Straight Line

Considered Formula; Motion Under Gravity

$$\text {i)}\quad v = u + gt$$ $$\qquad \text {Where}~~u \to Initial~velocity,\quad v \to Final~velocity \\\qquad a \to Acceleration~due~to~gravity \quad t \to Time$$ $$\text {ii)}\quad s =ut + \frac {1}{2}gt^2$$ $$\qquad \text {Where}~~s \to Distance~traveled,\quad u \to Initial~velocity, \\\qquad a \to Acceleration~due~to~gravity \quad t \to Time$$ $$\text {iii)}\quad v^2 = u^2 + 2gs$$ $$\qquad \text {Where}~~u \to Initial~velocity,\quad v \to Final~velocity, \\\qquad a \to Acceleration~due~to~gravity,\quad s \to Distance~traveled$$

Example 1

A stone of mass 2kg is released from a height of 20m above the ground. Find

a)    Velocity reached just before hit the ground

b)    Time taken to reach the ground

Solution

Data given

Distance (s) = 20 m

Accelaration due to gravity (g) = 10 m/s²

a)        Final velocity (v) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~v^2 = 2gs \quad \Rightarrow Initial~velocity~(u)= 0$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~v = \sqrt{2gs}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= \sqrt{2(10)(20)}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= \sqrt{400}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 20~m/s$$

          The final velocity = 20 m/s

b)        Time (t) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~v = gt$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~20 = 10t$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac {10t}{10}= \frac {20}{10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~t = 2~sec$$

          The time taken to reach the ground = 2seconds

Example 2

A stone is thrown vertically upward from the ground with a velocity of 30 m/s. find

a)    Maximum height reached

b)    Time taken to reach the maximum height

c)    Time taken to reach the ground again

d)    The velocity reached half-way to the maximum height

Solution

Data given

Initial velocity (u) = 30 m/s

Final velocity (v) = 0 m/s

Accelaration due to gravity (g) = 10 m/s²

a)        Distance (s) = ?
    For upward motion, g = -10 m/s²

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~v^2 = u^2 + 2gs$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0^2 = 30^2 + 2(-10)s$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 = 900 - 20s$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~20s = 900$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac {20s}{20}= \frac{900}{20}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~s= 45~m$$

          The maximum height reached = 45 m

b)        Time taken to reach maximum height (t₁) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~v = u + at$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 = 30 + 10t_1$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~10t_1 = 0 - 30$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~10t_1 = -30$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac {10t_1}{10}= \frac {-30}{10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~t_1 = 3~sec$$

          The time taken to reach maximum height = 3 seconds

c)        Time taken to reach ground again (t₂) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~t_2 = 2t_1$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 2 \times 3$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 6~sec$$

          The time taken to reach ground again = 6 seconds

d)        Final velocity (v) = ? $$\quad \text {Time (t)} = \Bigl ( \frac {3}{2} \Bigr ) = 1.5~sec$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~v = u + at$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 30 + (-10 \times 1.5)$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 30 - 15$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 15~m/s^2$$

          The velocity reached half-way to the maximum height = 15 m/s