Problem Analysis
We are given a projectile motion problem where a ball is thrown horizontally from the top of a building with an initial horizontal velocity of \(8.2 \, \text{m/s}\). At point \(P\), the ball is moving at an angle of \(60^\circ\) to the horizontal. We need to:
1.
Show that the vertical component of its velocity at point \(P\) is \(14.2 \, \text{m/s}\).
2.
Determine the vertical distance through which the ball has fallen.
Known Information:
- Initial horizontal velocity: \(v_{0x} = 8.2 \, \text{m/s}\)
- Angle at point \(P\): \(\theta = 60^\circ\)
- Acceleration due to gravity: \(g = 9.8 \, \text{m/s}^2\)
Solution:
#### Part (a)(i): Show that the vertical component of its velocity is \(14.2 \, \text{m/s}\).
At point \(P\), the ball's velocity vector can be resolved into horizontal and vertical components. Let the total velocity at point \(P\) be \(v_P\). The relationship between the components and the angle is given by trigonometry:
\[
\tan \theta = \frac{v_{Py}}{v_{Px}}
\]
where:
- \(v_{Px}\) is the horizontal component of velocity at point \(P\),
- \(v_{Py}\) is the vertical component of velocity at point \(P\).
Since there is no horizontal acceleration, the horizontal component of velocity remains constant:
\[
v_{Px} = v_{0x} = 8.2 \, \text{m/s}
\]
Given \(\theta = 60^\circ\):
\[
\tan 60^\circ = \sqrt{3}
\]
Using the tangent relationship:
\[
\tan 60^\circ = \frac{v_{Py}}{v_{Px}}
\]
\[
\sqrt{3} = \frac{v_{Py}}{8.2}
\]
Solving for \(v_{Py}\):
\[
v_{Py} = 8.2 \times \sqrt{3}
\]
\[
v_{Py} = 8.2 \times 1.732 \approx 14.2 \, \text{m/s}
\]
Thus, the vertical component of the velocity at point \(P\) is:
\[
\boxed{14.2 \, \text{m/s}}
\]
#### Part (a)(ii): Determine the vertical distance through which the ball has fallen.
To find the vertical distance fallen, we use the kinematic equation for uniformly accelerated motion:
\[
v_{Py}^2 = v_{0y}^2 + 2gh
\]
where:
- \(v_{Py}\) is the vertical component of velocity at point \(P\),
- \(v_{0y}\) is the initial vertical velocity,
- \(g\) is the acceleration due to gravity,
- \(h\) is the vertical distance fallen.
Since the ball is thrown horizontally, the initial vertical velocity is:
\[
v_{0y} = 0 \, \text{m/s}
\]
Substituting the known values:
\[
v_{Py}^2 = 0^2 + 2gh
\]
\[
(14.2)^2 = 2 \times 9.8 \times h
\]
\[
198.84 = 19.6h
\]
Solving for \(h\):
\[
h = \frac{198.84}{19.6}
\]
\[
h \approx 10.15 \, \text{m}
\]
Thus, the vertical distance through which the ball has fallen is:
\[
\boxed{10.15 \, \text{m}}
\]
Final Answers:
\[
\boxed{14.2 \, \text{m/s}}
\]
\[
\boxed{10.15 \, \text{m}}
\]
Parent Tip: Review the logic above to help your child master the concept of projectile motion worksheet.
