Question

A small rock dropped from a bridge hits the water after exactly 4 seconds. How high is the bridge?

Answer 1

Answer:

80 m approx

Explanation:

Use the formula S=ut+1/2at^2

where,

S=distance

U=initial velocity,where u is zero

t=time

a=acceleration,where acc is 9.8

S=ut+1/2at^2

=0x4+0.5x(9.8)x4^2

=0.5x(9.8)x16

=0.5x156.8

=78.4

approx 80 m

Answer 2

Given :

• Initial velocity, u = 0 m/s

• Acceleration due to gravity, g = 9.8 m/s²

• Time, t = 4 seconds

To find :

• Height of the bridge, h

According to the question,

Here we will use the formula,

➞ h = ut + ½ gt²

Where,

• h = Height
• u = Initial velocity
• g = Acceleration due to gravity
• t = Time

➞ Substituting the values,

➞ h = 0 × 4 + ½ × 9.8 × 4 × 4

➞ h = 0 + 9.8 × 2 × 4

➞ h = 0 + 78.4

➞ h = 78.4

So,the height is 78.4 m.

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More formulas :

Equation of motion for freely falling bodies :-

• v = u + gt

• h = ut + ½ gt²

• v² = u² + 2gs