Question

A stone is dropped from height of 50m on earth. At the same time , another stone is thrown vertically upwards from the ground with a velocity up from the ground of 50m/s . At what height from the ground will the two stones meet (g= -10m/s^2)

Answer 1

height = 45m

Step-by-step explanation:

`y = - (1)/(2) gt^2 + v_0t +x_0`

`y_1 = - (1)/(2) gt^2+50`

`y_2 = - (1)/(2) gt^2+50t`

`y_1 = y_2`

`50 = 50t\\t = 1`

`y_1(1) = - (1)/(2)g+50=-5+50=45`

Answer 2

At height of 45 meter from the ground two stones will meet each other.

Step-by-step explanation:

Let the height covered by the stone thrown up be x.

And time at which both will meet be t

A stone is dropped from height of 50 m on earth.

Distance from where the stone is dropped to point where it meets another stone= (50 -x) m

Initial velocity of the stone = u = 0 m/s

Acceleration due to gravity = g =
`10 m/s^2`

`h=ut+(1)/(2)gt^2` (Second equation of motion)

`(50 - x)= 0 m/s* t+(1)/(2)* 10 m/s^2 * t^2`

`(50 - x)= (1)/(2)* 10 m/s^2 * t^2`..(1)

A stone is tossed upward the ground .

Height of stone in 3.16 s = x

Initial velocity of the stone = u' = 50 m/s

Acceleration due to gravity = -g =
`-10 m/s^2`

`50-x=u't+(1)/(2)gt^2` (Second equation of motion)

`x = 50 m/s* t-(1)/(2)* (10 m/s^2) * t^2`...(2)

From (1) and (2)

t = 1 sec, x = 45 m

At height of 45 meter from the ground two stones will meet each other.