Question

a stone is thrown straight upward with an initial speed of 12m/s find the displacement and velocity after 2second

Answer 1

Answer: v' = 8 m/s² ;  S = 4 m.

Explanation:

Given,

Initial Speed (u) =  12 m/s

Final speed after 2 second (v') = ?

Displacement after 2 second (S) = ?

Let take g = -10 m/s² (As body is thrown vertically upward)

Velocity:-

Using  first Equation of motion, First we find the time taken to rich at highest point (For this v = 0)

v = u + gt

0 = 12 - 10t

10t = 12

t = 12/10

t = 1.2 second

Now, Stone will start falling downward.

We have to find velocity, after 2 second. But stone has already reached at highest point within 1.2 second. It means we have find velocity after (2-1.2) = 0.8 second from  point of falling down.

For this v' = ? , u = 0,  t = 0.8 sec , g = 10m/s²

v' = u + gt

v' = 0 + 10×0.8

v' = 8 m/s²

Displacement:-

First we find highest point from the ground.

Using second equation of motion,

S₁ =  ut + 1/2×gt²

Here, u = 12 , t = 1.2 second , g = -10 m/s²

S₁ = 12×1.2 - 1/2 × 10 1.2×1.2

S₁ = 14.4 - (5×1.2×1.2)

S₁ = 14.4 - (6× 1.2)

S₁ = 14.4 - 7.2

S₁ = 7.2 m

For  Downward motion,

S = S₂ , u = 0 , g = +10 m/s² , t = 0.8 sec

S₂ = ut + 1/2 × gt²

S₂ = 0 + 1/2 × 10 × 0.8 × 0.8

S₂ = 5 × 0.8 × 0.8

S₂ = 4 × 0.8

S₂ = 3.2 m

Now,

Displacement after 2 second,

S = S₁ - S₂

S = 7.2 - 3.2

S = 4 m.