Question

A piece of stone is thrown vertically upwards with a velocity 19.6 metre per second. How high will it rise? After how many seconds will it come back to the starting point? ​

Answer 1

Explanation:

Thus stone 1 can reach a maximum height of 19.6 meters in 2 seconds from ground when thrown at\[19.6m/s\]. In order for the second stone to reach the first stone at 19.6 meters, when it is thrown at \[9.8m/s\], it will take another 2 seconds to reach the point of maximum height

Answer 2

Answer:

Explanation:

Given, u=19.6m/s; s=h; v=0 [∵At the highest point, final velocity becomes zero]; a=g

From third equation of motion,

    v² = u² - 2gh

    0  = (19.6)² - 2 × 9.8 × h

    0  = 384.6 - 19.6 × h

    19.6 × h = 384.6

     h  = 384.6 / 19.6

     h = 19.62 m

 ∴The piece of stone will rise to a height of 19.62 m.

Now using first equation of motion,

    v = u + at

    0 = 19.6 - 9.8 × t

    9.8 × t = 19.6

     t = 19.6/9.8

     t = 2s

  Hence, After 2s it will reach its highest point.

  ∴ After 4s it will reach its starting point.

Please mark me brainliest :)