Question

An object drops from a building and strikes the ground with a speed of 31m/s.How tall is the building?

Answer 1

49.0m

Step-by-step explanation:

Given the following data;

U = 0m/s since it's a free fall

V = 31m/s

Also, since it's a free fall, a = g = 9.8m/s²

To find the height of the building, we would use the third equation of motion;

`V^(2) = U^(2) + 2aS`

Where;

• V represents the final velocity measured in meter per seconds.

• U represents the initial velocity measured in meter per seconds.

• a represents acceleration due to gravity measured in meters per seconds square.

• S represents the height measured in meters.

Substituting the values into the equation, we have;

`31^(2) = 0^(2) + 2*9.8*S`

`961 = 0 + 19.6S`

Simplifying, we have;

`S = \frac {961}{19.6}`

S = 49.03 ≈ 49.0

Therefore, the building is 49.0m tall.