Question

a stone is thrown vertically up with an initial velocity 40 m/s . taking g = 10 m/s ' find the maximum height reached by the stone . what is the net displacement covered by the body .

Answer 1

QUESTION ---- a stone is thrown vertically up with an initial velocity 40 m/s . taking g = 10 m/s ' find the maximum height reached by the stone . what is the net displacement covered by the body .


ANSWER
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Initial Velocity (u) = 40 m/sec
Acceleration (g) = 10 m/s^2
Final velocity (v) = 0 { It comes at rest on reaching maximum height}

Maximum height = ?

Using Third Equation of Motion,

v^2 = u^2 + 2as

0 = 40^2 + 2 × - 10 × s

-1600= -20 s

=> s = 80 m

So Maximum Height reached is equal to 80 m

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Net displacement covered by Stone = 0 m

It is because after throwing it up, It returns to the same position from where it was thrown.
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HOPE IT HELPS :);););););)

Answer 2

u= 40m/s
g= -10m/s²
v=0
h=?
.
v²-u²=2gh
0²-40²= 2×-10×h
-1600= -20×h
1600/20=h
80=h
•°• maximum height attained by the stone is 80m
As the body will fall back on the same place, therefore the net displacement is 0