A) It hits the ground at a horizontal distance 1.6m from the edge of the table
B) The speed with which it hits the ground is 4.0 m/s
C) Height of the table is 0.9 m
D) It hits the ground at an angle of \[60{}^\circ \] to the horizontal
Correct Answer: A
Solution :
\[\therefore \,\,\,R=u.T=4\times 0.4=1.6\,m\] \[{{V}_{y}}=gT=4\text{ }m/s\] \[\therefore \,\,\text{ }v=\,\,{{\sqrt{{{v}^{2}}_{x}+{{v}^{2}}}}_{y}}=4\sqrt{2}\text{ }m/s\]\[H=\frac{1}{2}\,\,g{{T}^{2}}\,\,=\,\frac{1}{2}\times 10\times {{(0.4)}^{2}}\,\,=\,\,0.8\,m\] \[~tan\text{ }\alpha =\,\,\frac{{{v}_{y}}}{{{v}_{x}}}=1\,\,\,\Rightarrow \text{ }\alpha =45{}^\circ \]You need to login to perform this action.
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