First slide

Projection Under uniform Acceleration

Question

A body is projected horizontally from the top of a tower of height h = 20 m with a speed u = 20 m/s . Then the angle, measured with horizontal, at which the body will strike the ground is

Moderate

A
450
B
600
C
300
D

$75^{0}$ $10\sqrt{5}$

Solution

Just before striking the ground ' horizontal component of the velocity of the body $v_{h} = 20 m / s$
Similarly vertical component of the velocity of the body $v_{v} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 20} m / s = 20 m / s .$
Therfore angle between velocity vector and horizontal = $\tan^{- 1} (\frac{v_{v}}{v_{h}}) = \tan^{- 1} (\frac{20}{20}) = 45^{0}$

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