Question

A ball is thrown horizontally from the top of a building 82 m high. The ball strikes the ground 53 m horizontally from the point of release. What is the speed of the ball just before it strikes the ground? Answer in units of m/s.

Answer 1

To find the speed of the ball just before it strikes the ground, we need to determine both the vertical and horizontal components of its velocity just before impact and then combine them to find the resultant speed.

1. **Vertical motion**: The ball falls a vertical distance of 82 m. We can use the following kinematic equation to find the time \( t \) it takes to fall this distance:

\[
h = \frac{1}{2} g t^2
\]

Where:
- \( h = 82 \, \text{m} \) (height of the building),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

Rearranging for \( t \):

\[
t^2 = \frac{2h}{g} \implies t^2 = \frac{2 \times 82}{9.81}
\]

Calculating \( t^2 \):

\[
t^2 = \frac{164}{9.81} \approx 16.7
\]

Now, taking the square root:

\[
t \approx \sqrt{16.7} \approx 4.08 \, \text{s}
\]

2. **Horizontal motion**: The horizontal distance traveled is 53 m in the same time \( t \). The horizontal velocity \( v_x \) can be given by:

\[
v_x = \frac{\text{horizontal distance}}{t} = \frac{53 \, \text{m}}{4.08 \, \text{s}} \approx 12.99 \, \text{m/s}
\]

3. **Vertical velocity just before impact**: The vertical velocity \( v_y \) can be calculated using:

\[
v_y = g t
\]

Calculating \( v_y \):

\[
v_y = 9.81 \times 4.08 \approx 40.05 \, \text{m/s}
\]

4. **Resultant velocity**: The resultant velocity \( v \) just before the ball strikes the ground can be found using the Pythagorean theorem:

\[
v = \sqrt{v_x^2 + v_y^2}
\]

Substituting the values of \( v_x \) and \( v_y \):

\[
v = \sqrt{(12.99)^2 + (40.05)^2}
\]
\[
v = \sqrt{168.74 + 1604.00} \approx \sqrt{1772.74} \approx 42.06 \, \text{m/s}
\]

Thus, the speed of the ball just before it strikes the ground is approximately **42.06 m/s**.