A) \[{{\tan }^{-1}}(1)\]
B) \[{{\tan }^{-1}}(5)\]
C) \[{{\tan }^{-1}}\left( \frac{1}{5} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{1}{5} \right)\]
Correct Answer: C
Solution :
Let h be height of plane from the ground, then from equation of motion, we have \[h=ut+\frac{1}{2}g{{t}^{2}}\] Since, initial velocity, u = 0 \[\therefore \] \[h=\frac{1}{2}g{{t}^{2}}\] Given, \[t=10\,s,\,g=10\ m/{{s}^{2}}\] \[\Rightarrow \] \[h=\frac{1}{2}\times 10\times {{(10)}^{2}}=500\,\,m\] Also, from equation \[{{v}^{2}}={{u}^{2}}+2gh\], we have for \[u=0,\,\,v=\sqrt{2\,gh}\] \[\therefore \] \[v=\sqrt{2\times 10\times 500}=100\,m/s\] Hence, \[\tan \theta =\frac{vertical\text{ }velocity}{horizontal\text{ }velocity}=\frac{100}{500}=\frac{1}{5}\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left( \frac{1}{5} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
