question_answer

From top of a building a ball is dropped while another is projected horizontally at the same time. (a) Which ball will strike the ground first? (b) Which ball will strike the ground with more speed.

Answer:

                As the horizontal velocity does not affect the vertical motion and initial vertical velocity of both the ball is zero, so both the balls will be reaching the ground simultaneously i.e. \[{{\text{t}}_{\text{1}}}={{\text{t}}_{\text{2}}}=\sqrt{2h/g}\] Velocity of the ball dropped vertically while reaching the ground is \[{{\upsilon }_{\text{1}}}=\sqrt{2gh}\] For the ball projected horizontally, the horizontal component velocity \[{{\upsilon }_{{{H}_{2}}}}=u\]and vertical component velocity, \[~{{\upsilon }_{{{v}_{2}}}}=\sqrt{2gh}\]. So, the total velocity of ball, \[{{\upsilon }_{2}}=\sqrt{\upsilon _{{{H}_{2}}}^{2}+\upsilon _{{{V}_{2}}}^{2}}=\sqrt{{{u}^{2}}+2gh}\]which is greater than\[{{\upsilon }_{1}}\]. It means the ball projected horizontally will strike the ground, with move speed than the ball dropped vertically downwards.