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
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