A) \[{{\text{a}}^{2}}y/{{\text{v}}_{0}}\]
B) \[{{\text{a}}^{2}}y/\sqrt{1+{{\left( \text{ay+}{{\text{v}}_{0}} \right)}^{2}}}\]
C) \[{{\text{a}}^{2}}y/\sqrt{1+{{\text{v}}_{0}}^{2}}\]
D) \[{{\text{a}}^{2}}{{v}_{0}}/\sqrt{1+{{\left( \text{2y+a} \right)}^{2}}}\]
Correct Answer: B
Solution :
| [b] Since velocity in vertical direction is constant, |
| \[\therefore \,\,\,\,{{\text{a}}_{\text{y}}}\text{=}\frac{\text{d}\,{{\text{v}}_{\text{y}}}}{\text{dt}}\text{=0}\] |
| The acceleration in horizontal direction, |
| \[{{a}_{x}}=\frac{d{{v}_{x}}}{dt}=\frac{d(a{{v}_{0}}t)}{dt}=a{{v}_{0}}\] |
| \[a=\sqrt{{{a}_{x}}^{2}+{{a}_{y}}^{2}}=\sqrt{{{\left( a{{v}_{0}} \right)}^{2}}+0}=a{{v}_{0}}\] |
| The total acceleration is \[a\,{{v}_{0}}\] and directed along horizontal direction. |
| Let \[\theta \] is the angle that the resultant velocity makes with horizontal, then |
| Normal acceleration \[{{a}_{n}}=a\,\,\sin \theta \] and tangential acceleration |
| \[{{a}_{t}}=a\cos \theta ,\text{ we have x}\,\text{=}\,\,\frac{a{{y}^{2}}}{2{{\text{v}}_{\text{0}}}}\]. |
| \[\text{or }\,\,\text{y=}\sqrt{\frac{2x{{v}_{0}}}{a}}\] |
| Differentiating both side of equation (iii) w.r.t. x, |
 |
| We get \[1=\frac{a}{2{{v}_{0}}}\times 2y\times \frac{dy}{dx}\] |
 |
| \[\text{or }\,\,\,\frac{dy}{dx}=\frac{{{v}_{0}}}{ay}=\tan \theta \] |
| Now \[{{a}_{x}}=a\sin \theta =a\,{{v}_{0}}\times \frac{\left( {{v}_{0}}/ay \right)}{\sqrt{1+{{\left( \frac{{{v}_{0}}}{ay} \right)}^{2}}}}\]\[=\frac{a{{v}_{0}}}{\sqrt{1+{{\left( \frac{ay}{{{v}_{0}}} \right)}^{2}}}}\]\[{{a}_{t}}=a\,\,\cos \theta =a\,{{v}_{0}}\times \frac{1}{\sqrt{1+{{\left( \frac{{{v}_{0}}}{ay} \right)}^{2}}}}=a\,{{v}_{0}}\]\[\frac{ay}{\sqrt{{{\left( ay \right)}^{2}}+{{v}_{0}}^{2}}}=\frac{{{q}^{2}}y}{\sqrt{1+{{\left( \frac{ay}{v{{ & }_{0}}} \right)}^{2}}}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
