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 bride is 9.8 m.

Explanation:

It is given that,

Initial speed of the stone, u = 4.9 m/s

It is thrown in vertically upward direction. It falls down in water after 2 seconds. We have to find the height of the bridge :

It can be calculated suing second equation of motion as :

$h=ut-\dfrac{1}{2}gt^2$

$h=4.9\ m/s\times 2\ s-\dfrac{1}{2}\times 9.8\ m/s^2\times (2\ s)^2$

h = -9.8 m

On neglecting negative sign, height of the bride is 9.8 m.

Answer 2

Answer:

9.8 m

Explanation:

Let height of bridge be equal to h

Noting that the kinematic expression

s = ut + ½at^2

where various symbols denote quantities as normally ascribed,

is a vector expression, taking origin at the bridge and taking up direction as positive; we see that acceleration due to gravity g = - 9.8 m/s^2.

Inserting various quantities we get  

-h = 4.9 × 2 - 0.5 × 9.8 × 2^2

Solving the equation we get

-h = 9.8 - 19.6

⇒ h = 9.8 m