Question

Question 1
A sparrow flying at 11.5 m/s wishes to have equal kinetic and gravitational potential energies. The height that will accomplish this is
Question 1 options:
a) 112.7 m
b) 0.6 m
c) 19.6 m
d) 6.7 m

Answer 1

To find the height at which the kinetic energy equals the gravitational potential energy for the sparrow, we can use the formulas for kinetic energy (KE) and gravitational potential energy (PE).

1. **Kinetic Energy (KE)** is given by the formula:
\[
KE = \frac{1}{2} mv^2
\]
where \(m\) is the mass of the sparrow and \(v\) is its velocity.

2. **Gravitational Potential Energy (PE)** is given by the formula:
\[
PE = mgh
\]
where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(h\) is the height.

Setting the two energies equal:
\[
\frac{1}{2} mv^2 = mgh
\]

Since mass \(m\) is present in both sides of the equation, we can cancel it (assuming the mass is not zero):
\[
\frac{1}{2} v^2 = gh
\]

Now, solving for \(h\):
\[
h = \frac{v^2}{2g}
\]

Substituting in the given values (\(v = 11.5 \, \text{m/s}\) and \(g = 9.81 \, \text{m/s}^2\)):
\[
h = \frac{(11.5)^2}{2 \cdot 9.81}
\]
\[
h = \frac{132.25}{19.62} \approx 6.73 \, \text{m}
\]

Thus, rounding to one decimal place, the height \(h\) is approximately \(6.7 \, \text{m}\).

So, the correct answer is **d) 6.7 m**.