Question

A stone is thrown vertically upward with the speed of 20m/s .How high will it goes before it begins to fall . (g=9.8 m/s)

Answer 1

hello there!!!!!

ATQ,

u= 20m/s. a=g = 9.8
v=0

as the stone is thrown upward so a= -g
Now , using the equation of motion,

v²- u² = 2aS

0² - (20)² = - 2 × 9.8 × S

- 400 = - 19.6 × S

S = 400/ 19.6 = 20.4m

Distance covered by the stone is 20.4 m



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