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