An aeroplane is flying in a horizontal direction with a velocity 600 km/h at height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance AB.

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Solution

From $h = \frac{1}{2} g t^{2}$
we have $t_{O B} = \sqrt{\frac{2 h_{O A}}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
Horizontal distance $A B = v t_{O B}$
$= (600 \times \frac{5}{18} m / s) (20 s)$
$= 3333.33 m = 3.33 k m$

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An aeroplane is flying in a horizontal direction with a velocity 600 km/h at height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance AB.

From h=12gt2we have tOB=√2hOAg=√2×19609.8=20sHorizontal distance AB=vtOB=(600×518m/s)(20s)=3333.33m=3.33km

From $h = \frac{1}{2} g t^{2}$
we have $t_{O B} = \sqrt{\frac{2 h_{O A}}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
Horizontal distance $A B = v t_{O B}$
$= (600 \times \frac{5}{18} m / s) (20 s)$
$= 3333.33 m = 3.33 k m$