On the earth, a stone(A) is thrown from a height in a direction parallel to the earths surface while another stone(B) is simultaneously dropped from the same height. Which stone would reach the ground first ?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both will reach at same time

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D Both will reach at same time
$s = u t + \frac{1}{2} a t^{2}$

For both A and B, $s = h, u = 0, a = g$
Hence, $t_{A} = t_{B}$

Suggest Corrections

Similar questions

Q. On Earth, two stones

A

and

B

are thrown vertically downwards from a same height at the same time. Which stone would reach the ground first? (Neglect air resistance)

Q. A stone is dropped from a certain height and at the same time another stone is thrown horizontally from the same height. Which one will reach the ground earlier? (neglect air resistance)

a stone dropped from a height h reaches at the earth surface in 1 second. If the same stone is taken to the moon and drop freely from a height h then in what time will it reach the ground?

Q. A stone is dropped from a height

h

. Another stone is thrown up simultaneously from the ground, which can reach to maximum height of

4 h

. The two stones will cross each other after time

Q. A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height of

4 h

. The two stones cross each other after time

Join BYJU'S Learning Program

On the earth, a stone(A) is thrown from a height in a direction parallel to the earths surface while another stone(B) is simultaneously dropped from the same height. Which stone would reach the ground first ?

The correct option is D Both will reach at same times=ut+12at2For both A and B, s=h,u=0,a=gHence, tA=tB

The correct option is D Both will reach at same time
$s = u t + \frac{1}{2} a t^{2}$

For both A and B, $s = h, u = 0, a = g$
Hence, $t_{A} = t_{B}$