- Analyze vertical and horizontal motion independently, noting zero initial vertical velocity and constant horizontal velocity.
- Solve the vertical motion equation \(h = \frac{1}{2}gt^2\) to find the time of flight \(t = 4 \, \text{s}\).
- Calculate horizontal displacement using \(d = v_{0x}t\) to find the range \(d = 120 \, \text{m}\).
- The final results are the time of flight \(\boxed{4 \, \text{s}}\) and the horizontal landing distance \(\boxed{120 \, \text{m}}\).
Explanation
- Problem Analysis and Motion Decomposition
The problem describes the motion of a projectile launched horizontally from the top of a building. We are given:
- The height of the building: \(h = 80 \, \text{m}\).
- The initial horizontal velocity: \(v_{0x} = 30 \, \text{m/s}\).
- The initial vertical velocity: \(v_{0y} = 0 \, \text{m/s}\) (since the ball is launched horizontally).
- The acceleration due to gravity: \(g = 10 \, \text{m/s}^2\), acting downwards.
Because there is no horizontal acceleration (\(a_x = 0\)), the horizontal velocity remains constant throughout the flight. The vertical motion is governed by constant downward acceleration due to gravity. We can analyze the vertical and horizontal components of the motion independently to find the time of flight and the horizontal distance.
- Calculating Time of Flight
To find the time it takes for the ball to reach the ground, we focus on the vertical motion. The vertical displacement \(h\) is given by the kinematic equation for motion with constant acceleration starting from rest:
\[h = v_{0y}t + \frac{1}{2}gt^2\]
Substituting the known values \(h = 80 \, \text{m}\), \(v_{0y} = 0 \, \text{m/s}\), and \(g = 10 \, \text{m/s}^2\):
\[80 = 0 \times t + \frac{1}{2} \times 10 \times t^2\]
\[80 = 5t^2\]
Solving for \(t^2\):
\[t^2 = \frac{80}{5} = 16\]
Taking the square root (choosing the positive root for time):
\[t = \sqrt{16} = 4 \, \text{s}\]
Thus, it takes \(4\) seconds for the ball to strike the ground.
- Calculating Horizontal Distance
With the time of flight determined, we can calculate the horizontal distance \(d\) from the base of the building. Since there is no horizontal acceleration, the horizontal distance is simply the product of the constant horizontal velocity and the time:
\[d = v_{0x} \times t\]
Substituting the values \(v_{0x} = 30 \, \text{m/s}\) and \(t = 4 \, \text{s}\):
\[d = 30 \, \text{m/s} \times 4 \, \text{s} = 120 \, \text{m}\]
The ball strikes the ground at a distance of \(120\) meters from the base of the building.
- Final Answer
In conclusion, the time elapsed before the ball strikes the ground is \(\boxed{4 \, \text{s}}\), and the horizontal distance from the base of the building where it strikes is \(\boxed{120 \, \text{m}}\).
Examples
Understanding projectile motion is essential in sports and engineering. For example, a basketball player releasing a ball at a certain angle and speed relies on these exact kinematic principles to predict the ball's trajectory and ensure it enters the hoop. Similarly, engineers designing safety features for vehicles or calculating the landing zone for an emergency supply drop use these formulas to ensure accurate results. Mastering these calculations allows you to model how objects move through the air under the influence of gravity.