A body dropped from a height of \(5 \mathrm{~m}\), reaches the ground in \(1 \mathrm{~s}\). If it takes \(2 \mathrm{~s}\) to reach the ground, find the height from which it is dropped.

Short Answer

Expert verified
Answer: The body must be dropped from a height of 20 meters to reach the ground in 2 seconds.

Step by step solution

01

Write down the given values

We are given: - The initial height \(h_1 = 5 \mathrm{~m}\) - The time taken to reach the ground for the first case \(t_1 = 1 \mathrm{~s}\) - The time taken to reach the ground for the second case \(t_2 = 2 \mathrm{~s}\)
02

Use the displacement formula for initial case

For the initial case, we use the formula of displacement for a falling object under uniform acceleration. When a body is dropped, the initial velocity \(u = 0\). The displacement formula is: $$ s = ut + \frac{1}{2}at^2 $$ Where - \(s\) is the displacement - \(u\) is the initial velocity - \(t\) is time taken to reach the ground - \(a\) is the acceleration due to gravity (approx \(9.81\,\text{m/s}^2\), but in most high school problems, \(a=10\,\text{m/s}^2\) is used) Using the given values for the initial case, we get: $$ 5 = 0 \cdot 1 + \frac{1}{2}a \cdot 1^2 $$
03

Calculate the acceleration due to gravity

By solving the equation from step 2, we can calculate the acceleration due to gravity \(a\): $$ a = 2 \cdot 5 = 10 \,\text{m/s}^2 $$
04

Use the displacement formula for the second case

Now we have the acceleration due to gravity, we can use the displacement formula from step 2 to find the height from which the body is dropped in the second case. Let this new height be \(h_2\): $$ h_2 = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot 2^2 $$
05

Calculate the new height

Now, we calculate the height \(h_2\) from the equation obtained in step 4: $$ h_2 = \frac{1}{2} \cdot 10 \cdot 4$$ $$ h_2 = 20 \,\text{m} $$ The body is dropped from a height of \(20 \,\text{m}\) to reach the ground in \(2 \,\text{s}\).

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