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A particle (A) is dropped from a height and another particle (B) is projected in horizontal direction with speed of 5 m/s from the same height then correct statement is : -
A
Particle (A) will reach at ground first with respect to particle (B)
B
Particle (B) will reach at ground first with respect to particle (A)
C
Both particles will reach at ground simultaneously
D
Both particles will reach at ground with same speed
Detailed Solution
For particle A (dropped from rest): $h = \dfrac{1}{2}gt_A^2 \Rightarrow t_A = \sqrt{\dfrac{2h}{g}}$
For particle B (projected horizontally), the initial vertical velocity is zero ($u_y = 0$). The horizontal velocity does not affect the vertical motion.
In the vertical direction, using $s = ut + \dfrac{1}{2}at^2$: $h = \dfrac{1}{2}gt_B^2 \Rightarrow t_B = \sqrt{\dfrac{2h}{g}}$
So $t_A = t_B$: both reach the ground simultaneously.
Their speeds on reaching the ground are different, since B also has a horizontal velocity of 5 m/s.
So both particles will reach the ground simultaneously.
For particle B (projected horizontally), the initial vertical velocity is zero ($u_y = 0$). The horizontal velocity does not affect the vertical motion.
In the vertical direction, using $s = ut + \dfrac{1}{2}at^2$: $h = \dfrac{1}{2}gt_B^2 \Rightarrow t_B = \sqrt{\dfrac{2h}{g}}$
So $t_A = t_B$: both reach the ground simultaneously.
Their speeds on reaching the ground are different, since B also has a horizontal velocity of 5 m/s.
So both particles will reach the ground simultaneously.