Question

A stone is dropped from the top of a 40m higher tower. calculate its speed after 2s. Also find the speed with which the stone strikes the ground.​

Answer 1

A stone is dropped from the top of a 40m higher tower. Therefore, the height or distance covered by the stone is 40 m.

We have to find the speed of the stone after 2 sec. Also, the speed of the stone with which the stone strikes the ground.

Initial speed of the stone is 0 m/s and acceleration due to gravity is 9.8 m/s².

Now,

Using the First Equation Of Motion i.e. v = u + at

Substitute the known values in the above formula,

→ v = 0 + 9.8(2)

→ v = 0 + 19.6

→ v = 19.6

Therefore, the final speed of the stone is 19.6 m/s.

Now, using the Third Equation Of Motion i.e. v² - u² = 2as

We have; u = 0 m/s, a = 9.8 m/s² and s = 40 m. Substitute the values,

→ v² - (0)² = 2(9.8)(40)

→ v² - 0 = 784

→ v = √784

→ v = 28

Therefore, the speed with which the stone strikes the ground is 28 m.

Answer 2

Question:

A stone is dropped from the edge of roof find the speed of the stone at 2 second after the stone was dropped.

Answer:

20 m/sec

Explained Answer:

Given:

Initial Velocity = 0m/s

Acceleration due to Gravity = 10m/s²

Time = 2s

Formula:

v = u + at

Here,

v = Final Velocity

u = Initial Velocity

a = Acceleration due to Gravity

t = Time

So, According to Question:

Asking:

Speed

v = u + at

v = 0 + 10 × 2

v = 20 m/sec

So, the speed of the stone at 2s after the stone was dropped is 20 m/sec.