First slide

Projection Under uniform Acceleration

Question

The friction of the air causes a vertical retardation equal to 10% of the acceleration due to gravity. Take g = 10 m/s². The maximum height and time to reach the maximum height will be decreased by

Difficult

A

9%, 9%
B

11%, 11%
C

9%, 10%
D

11%, 9%

Solution

$\begin{array}{l} h = \frac{u^{2} \sin^{2} θ}{2 g^{'}} = \frac{u^{2} \sin^{2} θ}{2 \times (\frac{11}{10} g)} \\ = \frac{10}{11} \times [\frac{u^{2} \sin^{2} θ}{2 g}] \\ = \frac{u^{2} \sin^{2} θ}{2 g} \times \frac{10}{11} \times 100 % \\ = 91 % of \frac{u^{2} \sin^{2} e}{2 g} \end{array}$
So, maximum height decreases by 9%
Further,
$\begin{array}{l} t = \frac{usin θ}{g^{'}} = \frac{usin θ}{(11 / 11) g} \\ = 91 % of \frac{usin θ}{g} \end{array}$
$∴$ time to reach the maximum height decreases by 9%.

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