Question

A stone is thrown up with an initial speed of 4.9 m/s from a bridge in vertically upward direction. it falls down in water after 2 sec the height of the bridge is​

Answer 1

Answer:

The height of the bridge is 9.8m.

Answer 2

Given :-

Initial velocity of the stone thrown up = 4.9 m/s

Time taken by the stone = 2 sec

To Find :-

The height of the bridge.

Analysis :-

Here we are given with the initial velocity, time taken and the acceleration due to gravity of the stone.

In order to find the height substitute the given values from the question using the second equation of motion.

Solution :-

We know that,

• s = Distance
• t = Time
• u = Initial velocity
• a = Acceleration

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} at^2}}$

Given that,

Initial velocity (u) = 4.9 m/s

Time (t) = 2 sec

Acceleration (a) = -9.8 m/s (Acceleration due to gravity, g)

Substituting their values,

$\sf s=4.9 \times 2 -\dfrac{1}{2} \times 9.8 \times 2^2$

$\sf s=9.8 - \dfrac{9.8 \times 4}{2}$

$\sf s=9.8 - \dfrac{39.2}{2}$

$\sf s=9.8 - 19.6$

$\sf s=-9.8 \ m$

Since height cannot be negative, s = 9.8 m

Therefore, the height of the bridge is 9.8 m.