A stone thrown vertically upward with 50 m/s velocity and g=10 m/s2. What will be the maximum height achieved by stone, net displacement and total distance covered by stone? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

The maximum height achieved by the stone is

125 m,

net displacement is zero and distance covered by the stone is

250 m .

Given,
Initial velocity,

u = 50 m / s

Acceleration,

a = 10 m / s^{2}

Final velocity,

v = 0 m / s^{}

(after achieving maximum height, the stone is at rest)
According to newton’s third equation of motion,

v^{2} = u^{2} + 2 as

\Rightarrow

0 = 5 0^{2} + 2 (- 10) s

\Rightarrow

s = 125 m

And the stone will fall to its initial position (on earth)
So total displacement will be

0

and total distance will be twice the maximum height.
Total distance,

= 2 s = 250 m

A stone thrown vertically upward with $50 m / s$ velocity and $g = 10 m / s^{2}$ . What will be the maximum height achieved by stone, net displacement and total distance covered by stone?