Question

an object is thrown directly downwards from a height of 60m with an initial velocity of 11m/s. what will be its velocity on impact

Answer 1

To find the final velocity of an object thrown directly downwards from a height of 60 meters with an initial velocity of 11 m/s, we can use the following kinematic equation:

\[
v^2 = u^2 + 2as
\]

where:
- \(v\) is the final velocity,
- \(u\) is the initial velocity (11 m/s, downward),
- \(a\) is the acceleration (due to gravity, approximately \(9.81 \, \text{m/s}^2\), downward),
- \(s\) is the displacement (60 m, downward).

Given:
- \(u = 11 \, \text{m/s}\)
- \(a = 9.81 \, \text{m/s}^2\)
- \(s = 60 \, \text{m}\)

Substituting these values into the equation:

\[
v^2 = (11 \, \text{m/s})^2 + 2(9.81 \, \text{m/s}^2)(60 \, \text{m})
\]

Calculating \( (11 \, \text{m/s})^2 \):

\[
(11 \, \text{m/s})^2 = 121 \, \text{m}^2/\text{s}^2
\]

Calculating \( 2(9.81 \, \text{m/s}^2)(60 \, \text{m}) \):

\[
2(9.81)(60) = 1177.2 \, \text{m}^2/\text{s}^2
\]

Now, substituting these values back into the equation:

\[
v^2 = 121 \, \text{m}^2/\text{s}^2 + 1177.2 \, \text{m}^2/\text{s}^2
\]
\[
v^2 = 1298.2 \, \text{m}^2/\text{s}^2
\]

Taking the square root to find \(v\):

\[
v = \sqrt{1298.2} \approx 36.0 \, \text{m/s}
\]

Thus, the final velocity of the object on impact will be approximately **36.0 m/s** (directed downwards).