First slide

Projection Under uniform Acceleration

Question

Two cliff of heights 120 m and 100.4 m are separated by a horizontal distance of 16 m. If a car has to reach from the first cliff to the second, the horizontal velocity of the car should be

Moderate

Solution

h = 19.6 m ; R = 16 m

$h = \frac{1}{2}g{t^2} \Rightarrow t = \sqrt {\frac{{2h}}{g}}$

$R = u\sqrt {\frac{{2h}}{g}} \Rightarrow 16 = u\sqrt {\frac{{2 \times 19.6}}{{9.8}}}$

$u = \frac{{16}}{2} = 8m{s^{ - 1}}$

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