A stone is dropped from a height of 45 meter . what will be the distance travelled by it during the last second of its motion ? take acceleration as g=9.8m/s^2

Question

A stone is dropped from a height of 45 meter . what will be the distance travelled by it during the last second of its motion ? take acceleration as g=9.8m/s^2

Answer 1

How long does it take to hit?

h(t) = 45-4.9t^2
when is h=0?
t = 3.03 seconds

Now, what is h(2.03)?

Answer 2

If g =10 then Ans is 25 m distance travelled in last second ..Now how to solve it we use formula h=1/2 gt^2,Hn =g(n-1/2)

Answer 3

To find the distance traveled by the stone during the last second of its motion, we need to determine the time it takes for the stone to fall from the height of 45 meters to just 1 meter above the ground (since it is the last second of motion).

We will calculate this using the basic kinematic equation of motion:

𝑠 = 𝑢𝑡 + 0.5𝑎𝑡²

Where:
- 𝑠 is the distance traveled
- 𝑢 is the initial velocity (which is 0 in this case since the stone is dropped)
- 𝑡 is the time elapsed
- 𝑎 is the acceleration due to gravity (𝑔 = 9.8 m/s²)

Since the initial velocity 𝑢 is 0, the equation simplifies to:

𝑠 = 0.5𝑎𝑡²

First, we need to find the time it takes for the stone to fall from a height of 45 meters to just 1 meter above the ground. We can use the second equation of motion:

𝑠 = 𝑢𝑡 + 0.5𝑎𝑡²

Since the final position (𝑠) is 1 meter above the ground, the equation becomes:

1 = 0 + 0.5(9.8)(𝑡²)

Rearranging the equation to solve for 𝑡:

0.5(9.8)(𝑡²) = 1

4.9(𝑡²) = 1

𝑡² = 1/4.9

𝑡 = √(1/4.9)

𝑡 ≈ 0.451 seconds

Now that we have the time it takes for the stone to fall from a height of 45 meters to just 1 meter above the ground, we can substitute this value back into the first equation to find the distance traveled:

𝑠 = 0.5𝑎𝑡²

𝑠 = 0.5(9.8)(0.451)²

𝑠 ≈ 1.048 meters

Therefore, the distance traveled by the stone during the last second of its motion is approximately 1.048 meters.