Question

2 marks
04) A ball is dropped from a height 'h'. The velocity of the
ball when it reaches the ground is 10 m/s.Find height 'h'(
take g = 10 m/s2)​

Answer 1

GIVEN :-

• A ball is dropped from a height 'h'.
• Final velocity (v) = 10m/s.
• g = 10m/s².

TO FIND :-

• Height (h).

SOLUTION :-

We have ,

• Initial velocity (u) = 0m/s [ball is dropped]
• Final velocity (v) = 10m/s²
• g = 10m/s²
• h = ?

By 3rd Kinematical equation,

★ v² = u² + 2gh

Putting values,

→ 10² = 0² + 2(10)(h)

→ 100 = 20h

→ h = 100/20

→ h = 5m

Hence , height from which ball is dropped is 5m.

MORE TO KNOW :-

1) 1st Kinematical equation :-

★ v = u + at

2) 2nd Kinematical equation :-

★ s = ut + (1/2)at²

Here ,

• v = Final velocity
• u = Initial velocity
• t = Time taken
• a = Acceleration
• s = Displacement

3) Kinematical equations are only applicable when acceleration is constant.

Answer 2

$\\\large\underline{ \underline{ \sf{ \red{Given:} }}}$

• Initial velocity = 0 m/s (u)
• Final velocity = 10 m/s (v)
• Acceleration = 10 m/s² (g)

$\\\large\underline{ \underline{ \sf{ \red{To \: Find:} }}}$

• Height (h) = ?

$\\\large\underline{ \underline{ \sf{ \red{Solution:} }}}$

We know,

Initial velocity = 0 m/s (u)

Final velocity = 10 m/s (v)

Acceleration = 10 m/s² (g)

Time taken = ? (t)

g = v-u/t

t = v-u/g

t = 10m/s - 0m/s / 10 m/s²

t = 10 m/s / 10 m/s²

t = 1 sec

Distance covered by the ball,

S= ut+1/2 gt²

s = (0) × 1 + 1/2 × 10 × 1

s = 0+5

s = 5m

$\boxed{h=5m}$

Hence, the height of the ball which is dropped = 5m.