Question

A particle A is projected from the ground with an initial velocity of 10 m/s at an angle of 60 degree with the ground. From what height should another particle B be projected horizontally with velocity 5m /s so that both the particles collide in ground at the same position,if both are projected simultaneously

Answer 1

Answer:

15m

Option ⇒ D

Explanation:

Correct Question:                       (With Attachment)

A particle A is projected from the ground with an initial velocity of 10 m/s at an angle 60° with horizontal. From what height h should another particle B be projected horizontally with velocity 5 m/s. so both the particle collide in the ground at point C if both are projected simultaneously (g=10m/s²)

See Attachment! It is also a part of question.

Options:

a)10m

b)30m

c)25m

d)15m

Solution:

Horizontal component of velocity of A is 10 cos 60°=5m/s

The two will collide if time taken for both A and B to reach C is same.

$\sf => t_1 = t_2\\\ From\;the\;second\;equation\;of\;motion,\\\ h = \dfrac{1}{2}gT^2B\\\ t_B=\sqrt{\dfrac{2h}{g}}\\\ t_A = \dfrac{2u\; sin\theta}{g}\\\ \dfrac{2u\; sin\theta}{g} = \sqrt{\dfrac{2h}{g}}\\\ h=\dfrac{2u^2\;sin^2\; \theta}{g}$

$\sf \dfrac{2.10^2 \bigg(\dfrac{\sqrt{3}}2\bigg)}{10}\\\ =15m\\Hence\;Answer\;is\;15m$