Question

A ball thrown vertically upwards with an initial velocity of 1.4m/s returns in 2s the total displacement of the ball is

Answer 1

Answer :-

→ Distance= 0.1 m .

→ Displacement = 0

To Find :-

→ Displacement of the ball .

Explanation :-

Given that :-

• Initial velocity (u) = 1.4 m/s

• Final velocity (v) = 0 , because it comes into rest when comes back .
• Time (t) = 2 seconds .
• displacement (s) = ?

Hence ,

Apply equation of motion for time displacement Equation .

$\implies \boxed{ \sf{ {v}^{2} = {u}^{2} + 2as \: }} \\ \\ \sf{ here \: acceleration \: due \: to \: gravity }\\ \sf{g }= \bf{ 9.8 \: \frac{m}{ {s}^{2} } } \\ \\ \implies \sf{ {0}^{2} = {(1.4)}^{2} - 2 \times 9.8 \times s \: } \\ \\ \implies \sf{19.6 \: \times s = 1.96} \\ \\ \implies \sf{ s = \frac{1.96}{19.6} } \\ \\ \implies \: \boxed{\sf{s = 0.1}}$

hence its distance is 0.1 m

Displacement is zeros because it returns to its initial position ,hence displacement is "0" .

Answer 2

Explanation:

Given -

u = 1.4 m/s (initial velocity)

v = 0 m/s (final velocity)

t = 2 seconds (time)

To Find -

Total Displacement of the ball

Now,

Here we use

s = (v + u/2)t

here,

v = final velocity

u = initial velocity

a = acceleration

s = distance travelled

t = time

» (0 + 1.4/2)2

» 0.7 × 2

» 1.4 m

Hence,

Distance travelled by ball of is 1.4 m.

And

As we know that :-

Displacement :-

Shortest distance between initial and final position.

so,

here, Displacement of ball is 0(Zero).

Some related formulas -

• v = u + at
• s = ut + 1/2at²
• v² = u² + 2as

here,

v = final velocity

u = initial velocity

a = acceleration

t = time

s = distance travelled