A ball is dropped from the roof of a tall building and simultaneously another ball is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your answer.

Explanation

Let `h` be the height of the tall building.
For dropped ball:
Let ``t_1`` be the time taken by the dropped ball to reach the ground.
Initial velocity , u=0, Acceleration , a= + g
Distance travelled, s=h, Time of travel , `t=t_1`
From the equation of motion, `s=ut+1/2at^2`
we can write
`h=0timest_1+1/2times g timest_1^2implies h=1/2 g t _1^2impliest_1^2=(2h)/g`
(or) `t_1=sqrt((2h)/g)` ____(1)
For horizontally projected ball:
If the ball is thrown horizontally then its initial velocity along vertical direction is zero and in this case let ``t_2`` be the time taken by the ball to reach the ground.
Again from the equation of motion,
`s=ut+1/2at^2` we write,
`impliesh=0timest_2+1/2 g times t_2^2`
`impliesh=1/2g t _2^2`,
`implies t_2^2=(2h)/g`or`t_2=sqrt((2h)/g)`______(2)
From equation s (1) and (2) `t_1=t_2`
i.e., both the balls reach the ground in the same time.