Question

a stone is thrown vertically upward with a speed of 20 m per second how high will it go before it begins to fall

Answer 1

ʜᴇʟʟᴏ ᴍᴀᴛᴇ!

Here, u ( initial velocity ) = 20 m/s
v = 0
a = - 10 m/s²

v² - u² = 2as [ Third equation of motion ]

(0)² - (20)² = 2(-10)(s)

0 - 400 = -20s

- 400 / - 20 = s

s ( distance ) = 20 m

Hence, ball will go 20 m high when thrown vertically up by 20 m/s velocity.

Hope it helps

Answer 2

initial velocity (u) = 20m/s
final velocity (v) = 0
acceleration (a) = -9.8 m/s² ,which can be taken as -10m/s²

As third equation of motion is,

v² - u² = 2as

Here,
'v' is final velocity, 'u' is initial velocity, 'a' is the acceleration and 's' is distance.

=> (0)² - (20)² = 2(-10)(s)

[ we can use the identity :
a² - b² = (a+b)(a-b) ]

=> (0 + 20)(0 - 20) = -20s

=> 20 × -20 = -20s

=> -400 = -20s

=> -400/-20 = s

=> 20 = s

Hence, the stone will cover (s) distance 20m before it begins to fall.