Question

a stone is thrown vertically upward with a speed of 20 metre per second. How will it go before it begins to fall? The value of g is 9.8 metre per second square

Answer 1

Velocity at maximum height (v) = 0
Acceleration = -g
(-ve sign because it’s in the direction opposite to motion of stone)

v² - u² = 2aS

S = v² - u² / (2a)
= [0 - (20 m/s)²] / [2 × (-9.8 m/s²)]
= 20.40 m

It will reach a height of 20.40 m before it begins to fall.

hope helps you ❤️❤️❤️❤️

love you all
any doubt comments below
follow me

Answer 2

HEYA MATE, HERE IS UR ANSWER

Initial velocity (u)=20 m/sec

Final velocity ( v ) = 0 m/sec [as the stone stops for some time ]

Acceleration due to gravity=-9.8 m/sec^2
[stone thrown upwards]

By Third equation of motion

v^2=u^2+2as

0^2=20^2+2×-9.8 s

0=400-19.6s

0-400=-19.6s

$\frac {-400}{-19.6}$=s

20.4 m (approx )=s


So the distance travelled =20.4m


$\mathbb{Be\:Brainly}$