Question

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Answer 1

Given:-

• Time of flight = t = 3 seconds
• g = 9.8 m/s²

To find:-

• Height from which the ball was released = h
• The velocity with which the ball will strike the ground = v

Answer:-

Answer (a):-

s = ut + (1/2)at²

▪Given: t = 3 seconds

▪Since the ball is released from rest,

initial velocity = u = 0 m/s

▪Sign convention: 's' will be (-h) since

it is in downward direction. 'a' will

be (-g) since it is acting in

downward direction.

Putting this in the above formula,

→ -h = (0 × 3) + [1/2 × (-g) × 3 × 3]

→ -h = 1/2 × (-9.8) × 3 × 3

→ h = 1/2 × 9.8 × 3 × 3

→ h = 44.1 m Ans.

Answer (b):-

v² = u² + 2as

▪Since the ball is released from rest,

initial velocity = u = 0 m/s

▪Sign convention: 's' will be (-h) =

(-44.1) since it is in downward

direction. 'a' will be (-g) since it is

acting in downward direction.

Putting this in the formula,

→ v² = 0² + [2 × (-9.8) × (-44.1)]

→ v² = 2 × 9.8 × 44.1

→ v² = 864.36

→ v = ±29.4 m/s

But the final velocity (v) will be negative, since it is in downward direction.

→ v = -29.4 m/s Ans.

Answer 2

Answer :-

Given :-

• Time taken, t = 3 sec
• Acceleration due to gravity, g = 9.8 m/s²
• Initial Velocity, u = 0 ( because the object is released )

To Find :-

• Height, h
• Final velocity, v

Solution :-

Calculating Height :-

• u = 0 m/s
• t = 3 s
• a = g = 9.8 m/s²

Substituting the value in 2nd equation of motion :-

→ s = ut + ½ at²

→ s = 0 + ½ × 9.8 × 3²

→ s = 4.9 × 9

→ s = 44.1 m

Height from which it was released is 44.1 m

Calculating final velocity :-

• u = 0 m/s
• t = 3 s
• a = g = 9.8 m/s²

Substituting the value in 1st equation of motion :-

→ v = u + at

→ v = 0 + 9.8 × 3

→ v = 14.7

Velocity with which it strikes ground is 14.7 m/s