A) 12.5 m/s
B) 25 m/s
C) 31.25 m/s
D) 40 m/s
Correct Answer: D
Solution :
Let body be projected at an angle \[\theta \] with the horizontal at velocity u, then its time of flight is \[T=\frac{2u\,\,\sin \theta }{g}\] ? (i) Also horizontal range\[R=u\cos \theta \times t\] \[R=\frac{u\cos \theta \times 2u\sin \theta }{g}\] ?. (ii) Given, \[T=5\,\,s,\] \[R=200\,\,m\] \[\therefore \]From Eqs. (i) and (ii), we get \[5=\frac{2u\,\,\sin \theta }{g}\] ... (iii) \[200=\frac{2{{u}^{2}}\sin \theta cos\theta }{g}\] ? (iv) Dividing Eq. (iv) by Eq. (iii), we get \[\frac{200}{5}=u\cos \theta =40\,\,m/s\]You need to login to perform this action.
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