Question

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 strikes the ground with more speed?

Answer 1

$Initially$ $Initially$

$V_{AY}=0$ $V_{B}Y=0$

$h=0t_A+\frac{1}{2}g{t}_A^2$ $\to h=o{t}_B+\frac{1}{2}g{t}_B^2$

$t_A=\sqrt{\frac{2h}{g}}$ $t_B=\sqrt{\frac{2h}{g}}$

$\to$ They reach at same time, it depends on vertical velocity.

For A,

Finally,

$V_A^{\prime }=\sqrt{V_{AX}^{\prime 2}+V_{AY}^{\prime 2}}$

$=\sqrt{V^2+V_{AY}^{\prime 2}}$

For B,

$V_B^{\prime }=\sqrt{V_{BX}^{\prime 2}+V_{BY}^{\prime 2}}$

$=\sqrt{0+V_{BY}^{\prime 2}}$

$V_{AY}^{\prime }=0+gt,V_{BY}^{\prime }=0+gt$

$\Rightarrow V_{AY}^{\prime }=V_{BY}^{\prime }$

$\Rightarrow V_A^{\prime }>V_B^{\prime }$

$\Rightarrow$ Ball A will strike with more velocity