question_answer

A ball is rolled off the edge of a horizontal table at a speed of 4 m/second. It hits the ground after 0.4 second. Which statement given below is true

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 \[\operatorname{h}=ut=4\,\,\times \,\,0.4=1.6m\]