Question

Due to the presence of air resistance, if a body dropped from a height of 20 m reaches the ground with a speed of 18ms$^{-1}$, then the time taken by the body to reach the ground is nearly

• A. 1.8 s
• B. 2.2 s
• C. 2 s
• D. 2.5 s

Hint:

When a problem involves non-uniform acceleration (like air resistance) but doesn't provide the force law, check if a simple kinematic equation can be used to find an approximate answer. The equation $s = \left(\frac{u+v}{2}\right)t$ is valid for constant acceleration, but it can also be interpreted as displacement = average velocity $\times$ time, which can be a reasonable approximation.

Answer 1

The Correct Option is B

This problem involves motion with non-constant acceleration due to air resistance. However, we are not given the law for air resistance. The options suggest that we can find an answer using the basic kinematic equations, which implies we should use an average acceleration or relate the average velocity.
Let's use the kinematic equation that relates displacement, initial velocity, final velocity, and time, assuming a constant (average) acceleration:
$h = \left(\frac{u+v}{2}\right)t$.
Here, the displacement is $h=20$ m.
The initial velocity is $u=0$ m/s (since it's dropped).
The final velocity is $v=18$ m/s.
We need to find the time $t$.
$20 = \left(\frac{0+18}{2}\right)t$.
$20 = (9)t$.
$t = \frac{20}{9}$ s.
$t \approx 2.222...$ s.
This value is approximately 2.2 s, which matches option (B). This approach assumes the acceleration is constant, which isn't true with air resistance, but it's the most reasonable interpretation given the limited information and options.