Question

An object is dropped from the top of a building and strikes the ground 3.0 s later. How tall is the building? *

Answer 1

Given:-

• Initial velocity ,u = 0m/s

• Time taken ,t = 3

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

To Find :-

• Height of the Building ,h

Solution:-

Using 2nd Equation of Motion we get

• h = ut + 1/2at²

Substitute the value we get

→ h = 0×3 + 1/2×9.8 ×3²

→ h = 0 + 4.9 × 9

→ h = 4.9 × 9

→ h = 44.1 m

Therefore, the height of the building is 44.1 metres.

Additional Information!!

Some equation of motion are

• v = u +at

• s = ut +1/2at².

• v² = u² + 2as

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement or Distance

t is the time taken

Answer 2

Answer:

h=gt2,

h=10*3*3=90m

Explanation:

here u =o when an object is thrown from height it's initial velocity =0.