From the top of a tower of height 78.4m two stones are projected horizontally with 10m/s and 20m/s …..

Q. From the top of a tower of height 78.4 m two stones are projected horizontally with 10 m/s and 20 m/s in opposite directions. On reaching the ground, their separation is

(a)120 m

(b)100 m

(c) 200 m

(d) 150 m

Ans: (a)

Sol:

$ \displaystyle x_1 = v_1 t_1 $

$\displaystyle = v_1 \sqrt{\frac{2h}{g}}$

$ \displaystyle x_1 = 10 \sqrt{\frac{2\times 78.4}{9.8}} = 40 m$

Similarly ,

$ \displaystyle x_2 = 20 \sqrt{\frac{2\times 78.4}{9.8}} = 80 m $

As two stones are moving in opposite direction.

Hence separation between them is

= 40 + 80 = 120 m

Author: Rajesh Jha

QuantumStudy.com

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