Take \[g=10\,\,m/{{s}^{2}}\]
A) 8s
B) 6.28 s
C) 5.6s
D) 7.07s
Correct Answer: D
Solution :
Consider the point of projection as origin and horizontal direction as positive X-axis and vertical upward direction as positive Y-axis. For motion along Y-axis, from \[y={{u}_{y}}t+\frac{{{a}_{y}}{{t}^{2}}}{2}.\] For particle to reach ground, y = - 200 m \[\therefore \] \[-200=(10\sin {{45}^{o}})t-\frac{g{{t}^{2}}}{2}\] The two roots of this quadratic equation are 7.07 sand-5.66s. Negative time is not possible, so required time is 7.07s.
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