NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Physics
- Chapter
- Motion in Two Dimensions
NTA Abhyas 2020 · Physics · Motion in Two Dimensions
A particle is thrown over a triangle from one end of a horizontal...
A particle is thrown over a triangle from one end of a horizontal base and after grazing the vertex falls on the other end of the base. If and be the base angles and be the angle of projection. Find the value of .
Options
- A\text\tan \alpha + \text\tan \beta
- B\text\sin \alpha + \text\sin \beta
- C\text\tan \alpha + \text\sin \beta
- D\text\cos \alpha + \text\cos \alpha
Correct answer
A. \text\tan \alpha + \text\tan \beta
Step-by-step solution
The situation is shown in diagram
\text\tan α + \text\tan β = \dfrac{\text{y}}{\text{x}} + \dfrac{\text{y}}{\text{R} - \text{x}}
\text\tan α + \text\tan β = \dfrac{\text{yR}}{\text{x} \left(\text{R} - \text{x}\right)} ...(i)
Equation of trajectory is \text{y} = \text{x tan} θ \left[1 - \dfrac{\text{x}}{\text{R}}\right]<br /><br />or, \text\tanθ = \dfrac{\text{yR}}{\text{x} \left(\text{R} - \text{x}\right)} ...(ii)<br /><br />From Eqs. (i) and (ii), we have<br /><br /> \text\tanθ = \text\tanα + \text\tanβ
\text\tan α + \text\tan β = \dfrac{\text{y}}{\text{x}} + \dfrac{\text{y}}{\text{R} - \text{x}}
\text\tan α + \text\tan β = \dfrac{\text{yR}}{\text{x} \left(\text{R} - \text{x}\right)} ...(i)
Equation of trajectory is \text{y} = \text{x tan} θ \left[1 - \dfrac{\text{x}}{\text{R}}\right]<br /><br />or, \text\tanθ = \dfrac{\text{yR}}{\text{x} \left(\text{R} - \text{x}\right)} ...(ii)<br /><br />From Eqs. (i) and (ii), we have<br /><br /> \text\tanθ = \text\tanα + \text\tanβ