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)
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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
hope it helps ⭐⭐⭐
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
hope it helps ⭐⭐⭐
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