Question

An object is thrown along a direction inclined at an
angle of 45° with the horizontal
direction. The horizontal
range of the particle is :
(A) Four times the vertical height
(B) Thrice the vertical height
(C) Twice the vertical height
(D) Equal to vertical height​

Answer 1

Option A is the right answer.

GIVEN:

Angle with respect to horizontal, θ = 45°

TO FIND:

• Horizontal range, R
• Vertical height, H

FORMULAE:

• sin 90° = 1

• sin 45° = 1/√2

• Horizontal range, R = (u² sin 2θ) /g

• Vertical height, H = (u² sin²θ ) /2g

Where,

• u is the velocity with which the object is thrown.
• g is the acceleration due to gravity.

SOLUTION:

STEP 1: TO FIND HORIZONTAL RANGE:

R = (u² sin 2θ) /g

$= \frac{ {u}^{2} \sin(2 \times 45) }{g} \\ \\ = \frac{ {u}^{2} \sin(90) }{g} \\ \\ = \frac{ {u}^{2}(1) }{g} \\ \\ Horizontal \: range = \frac{ {u}^{2} }{g}$

STEP 2: TO FIND VERTICAL HEIGHT:

H = (u² sin²θ ) /2g

$= \frac{ {u}^{2} { \sin(45) }^{2} }{2g} \\ \\ = \frac{ {u}^{2} { (\frac{1}{ \sqrt{2} }) }^{2} }{2g} \\ \\ = \frac{ {u}^{2}( \frac{1}{2} ) }{2g} \\ \\ Vertical \: height= \frac{ {u}^{2} }{4g}$

STEP 3: COMPARE RANGE AND HEIGHT:

$\frac{range}{height} = \frac{ \frac{ {u}^{2} }{g} }{ \frac{ {u}^{2} }{4g} } \\ \\ = \frac{ {u}^{2} }{g} \times \frac{4g}{ {u}^{2} } \\ \\ = \frac{4g}{g} \\ \\ \frac{range}{height} = 4 \\ \\ horizontal \: range = 4 \times vertical \: height$

ANSWER:

The horizontal range of the particle is four times the vertical height.

HORIZONTAL RANGE:

• It is the maximum distance to which the object projected can reach the ground.

• It is the distance between point of projection and the point where it is landed.

VERTCAL HEIGHT:

• It is the maximum height through which the object can travel high in air.

• At the highest point , the vertical velocity of object is zero.