Question

3
After
A particle is projected with a velocity of 30 m/s at an angle with the horizontal, where 0 = tan
4
1 second, direction of motion of the particle makes an angle a with the horizontal then a is given by
1
1
(2) tana =
2
(1) tana =
3
(4) tana = 1
CHES
2.
(3) tana​

Answer 1

Correct Question:

A particle is projected with a velocity of 30 m/s at an angle θ with the horizontal, where $\sf \theta = tan^{-1}\bigg(\dfrac{3}{4}\bigg)$. After 1 second, direction of motion of the particle makes an angle α with the horizontal then α is given by

$\sf (1) \ \alpha = tan^{-1}(1)$

$\sf (2) \ \alpha = tan^{-1}(2)$

$\sf (3) \ \alpha = tan^{-1}\bigg(\dfrac{1}{3}\bigg)$

$\sf (4) \ \alpha = tan^{-1}\bigg(\dfrac{1}{3}\bigg)$

Given:

Initial velocity of projection (u) = 30 m/s

Angle of projection $\sf (\theta) = tan^{-1}\bigg(\dfrac{3}{4}\bigg)$

To Find:

Angle made by particle with horizontal after 1 second (α)

Answer:

Let the velocity of particle after 1 second be 'v'

As there is no horizontal acceleration, horizontal component of velocity does not change.

$\rm \leadsto u \: cos \theta = v \: sin \alpha \: \: \: \: \: \: \: ...eq_1$

As, $\rm tan \theta = \dfrac{3}{4}$

So,

$\rm sin \theta = \dfrac{3}{5}$

$\rm cos \theta = \frac{4}{5}$

By using first equation of kinematics for the vertical component of velocity, we get:

$\rm \leadsto v \: sin \alpha = u \: sin \theta - gt \\ \\ \rm \leadsto v \: sin \alpha =30 \times \dfrac{3}{5} - 10 \times 1 \\ \\ \rm \leadsto v \: sin \alpha =6 \times 3 - 10 \\ \\ \rm \leadsto v \: sin \alpha =8 \: \: \: \: \: \: \: ...eq_2$

Dividing $\sf eq_2$ by $\sf eq_1$:

$\rm \leadsto \dfrac{ \cancel{v} \: sin \alpha }{\cancel{v} \: cos \alpha } = \dfrac{8}{u \: cos \theta} \\ \\ \rm \leadsto \dfrac{ sin \alpha }{ cos \alpha } = \dfrac{8}{30 \times \dfrac{4}{5} } \\ \\ \rm \leadsto tan \alpha = \dfrac{8}{6 \times 4 } \\ \\ \rm \leadsto tan \alpha = \dfrac{8}{24} \\ \\ \rm \leadsto tan \alpha = \dfrac{1}{3} \\ \\ \rm \leadsto \alpha = tan^{-1}\bigg(\dfrac{1}{3}\bigg)$

$\therefore$ Angle made by particle with horizontal after 1 second (α) = $\sf tan^{-1}\bigg(\dfrac{1}{3}\bigg)$