Question

An object is thrown straight from the top of the building at a speed of 20m/s. It hits the ground with a speed of 40m/s. How long was the object in the air?

Answer 1

$\huge\frak{AnswEr :}$

Given that a object is thrown straight down from the top of a building at a speed of 20 m/s it hits the ground with a speed of 40m/s.

That is,

• Initial velocity, u = 20 m/s
• Final velocity, v = 40 m/s
• Acceleration due to gravity, g = 9.8m/s²

We've to calculate the total time when the object was in the air. It can be calculated by using the equations of motion.

You can easily derive the equations of motion by algebraic, calculus & by graphical methods.

By using the 1st equation of motion :

$\dag \: \large{ \boxed{ \bf{ v = u + at}}}$

where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration [Acceleration due to gravity]
• t = Time taken

Substituting the values :

$\longrightarrow\sf 40 = 20 + 9.8 \times t$

$\longrightarrow\sf 9.8 t = 40 - 20$

$\longrightarrow\sf 9.8t = 20$

$\longrightarrow\sf t = \dfrac{20}{9.8}$

$\longrightarrow \large{ \boxed{\sf{ t = 2.04 \: seconds}}}$

$\therefore$ The object was 2.04 sec in the air.