First slide

Projection Under uniform Acceleration

Question

Form the top of a tower of height 78.4 m, two stones are projected horizontally with 20 ms^-1 and 30 ms^-1 in opposite directions. On reaching the ground. their separation is

Easy

A

100 m
B

200 m
C

50 m
D

400 m

Solution

Separation between them $R_{1} + R_{2}$ .
$U_{1} \sqrt{\frac{2 h}{g}} + U_{2} \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 h}{g}} (20 + 30) = 200 m$ .

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