A) It hits the ground at a horizontal distance m from the edge of the table
B) The speed with which it hits the ground is 4.0 m/second
C) Height of the table is 0.8 m
D) It hits the ground at an angle of \[60{}^\circ \] to the horizontal
Correct Answer: C
Solution :
(a, c) Vertical component of velocity of ball at point P \[{{\nu }_{v}}=0+gt=10\,\,\times \,\,0.4\,\,=\,\,4 m/s\] Horizontal component of velocity = initial velocity \[\Rightarrow \,\,\,\,\,\,\,\,\nu H\,=4\,m/s\] So the speed with which it hits the ground \[\nu =\sqrt{\nu _{H}^{2}+\nu _{V}^{2}}\,\,=\,\,4\sqrt{2}\,\,m/s\] and \[\tan \,\,\theta =\frac{{{\nu }_{V}}}{{{\nu }_{H}}}=\frac{4}{4}=1\,\,\,\,\Rightarrow \,\,\,\,\,\theta =45{}^\circ \] It means the ball hits the ground at an angle of \[45{}^\circ \] to the horizontal. Height of the table \[\operatorname{h} =\frac{1}{2} g{{t}^{2}}=\frac{1}{2}\times 10 \times {{\left( 0.4 \right)}^{2}}= 0.8 m\] Horizontal distance travelled by the ball from the edge of table \[\operatorname{h}=ut=4\,\,\times \,\,0.4=1.6m\]You need to login to perform this action.
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