Question

A particle is projected with speed 25m/s frum (0 0) .It passes through (25 85) then find (i) angle of projection cii) Range

Answer 1

AnswEr :

• Angle of Projection = 1.28°

• Range = 34.37 units

Explanation :

Given,

• Initial Velocity (u) = 25 m/s

• The particle is initially projected at (0,0) and subsequently passes (25,85)

To finD

Angle of Projection and Range

Angle Of Projection

(Refer to the attachment)

The particle passes through the coordinates (25,85). In other words,the particle has been displaced for 25 units horizontally (X axis) and 85 units vertically (Y axis)

Rearranging the position vectors (as shown in the attachment) would form a triangle

From Trigonometry,

$\tan( \alpha ) = \dfrac{85}{25} \\ \\ \longrightarrow \: \tan( \alpha ) = \dfrac{17}{5} \\ \\ \longrightarrow \: \boxed{ \boxed{ \alpha = { \tan }^{ - 1} ( \dfrac{17}{5} )}}$

The angle of projection would be the tangential inverse of 17/5

$\rule{300}{1}$

Range

Suppose r is the range of the above projectile

$\huge{ \boxed{ \boxed{ \rm R = \dfrac{u {}^{2} \sin(2 \alpha ) }{g} }}}$

Thus,

$\dashrightarrow \: \rm \: R = \dfrac{ {25}^{2} \times \sin(2.56) }{10} \\ \\ \dashrightarrow \: \rm \: R = \dfrac{625 \times 0.55}{10} \\ \\ \dashrightarrow \: \rm \: R = 62.5 \times 0.55 \\ \\ \dashrightarrow \: \boxed{ \boxed{ \rm R = 34.375 \: units}}$

NoTE

• tan^-1 (17/5) = 1.28

• sin(2.56) = 0.55

$\rule{300}{1}$

$\rule{300}{1}$