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

Answer:

height = 45m

Explanation:

$y = - \frac{1}{2} gt^2 + v_0t +x_0$

$y_1 = - \frac{1}{2} gt^2+50$

$y_2 = - \frac{1}{2} gt^2+50t$

$y_1 = y_2$

$50 = 50t\\t = 1$

$y_1(1) = - \frac{1}{2}g+50=-5+50=45$

Answer 2

Answer:

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

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 = $10 m/s^2$

$h=ut+\frac{1}{2}gt^2$ (Second equation of motion)

$(50 - x)= 0 m/s\times t+\frac{1}{2}\times 10 m/s^2 \times t^2$

$(50 - x)= \frac{1}{2}\times 10 m/s^2 \times t^2$..(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 = $-10 m/s^2$

$50-x=u't+\frac{1}{2}gt^2$ (Second equation of motion)

$x = 50 m/s\times t-\frac{1}{2}\times (10 m/s^2) \times t^2$...(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.