A ball is dropped off a high cliff, and 2 seconds later another ball is thrown vertically downward with an initial speed of 30 m/s. How long will it take the second ball to overtak... A ball is dropped off a high cliff, and 2 seconds later another ball is thrown vertically downward with an initial speed of 30 m/s. How long will it take the second ball to overtake the first?
Understand the Problem
The question is asking for the time it takes for a second ball, thrown downward with an initial speed, to catch up to a first ball that was dropped from a cliff two seconds earlier, involving concepts in kinematics.
Answer
The second ball takes approximately $1.89 \, \text{s}$ to overtake the first ball.
Answer for screen readers
The second ball will take approximately $1.89 , \text{s}$ to overtake the first ball.
Steps to Solve
- Establish the motion equations for both balls
For the first ball (dropped from rest), the distance it falls after $t_1$ seconds can be described by the equation: $$ d_1 = \frac{1}{2} g t_1^2 $$ where $g$ is the acceleration due to gravity ($9.81 , \text{m/s}^2$).
For the second ball, which is thrown downward after a delay of 2 seconds, the time it travels will be $t_2 = t_1 - 2$. Its motion can be described by: $$ d_2 = v_0 t_2 + \frac{1}{2} g t_2^2 $$ where $v_0 = 30 , \text{m/s}$.
- Set the time variables
We know the first ball has been falling for $t_1$ seconds, while the second ball has been falling for $t_2 = t_1 - 2$ seconds. We need to set both distances equal to each other at the moment they meet: $$ d_1 = d_2 $$
- Substituting distances into the equation
Substituting the expressions for $d_1$ and $d_2$ gives: $$ \frac{1}{2} g t_1^2 = v_0 (t_1 - 2) + \frac{1}{2} g (t_1 - 2)^2 $$
- Simplifying the equation
Substituting values for $g$ and $v_0$: $$ \frac{1}{2} \times 9.81 t_1^2 = 30 (t_1 - 2) + \frac{1}{2} \times 9.81 (t_1 - 2)^2 $$
Expanding the terms on the right side: $$ \frac{1}{2} \times 9.81 t_1^2 = 30 t_1 - 60 + \frac{1}{2} \times 9.81 (t_1^2 - 4t_1 + 4) $$
- Combining like terms
Combining terms: $$ \frac{1}{2} \times 9.81 t_1^2 = 30 t_1 - 60 + \frac{1}{2} \times 9.81 t_1^2 - 2 \times 9.81 t_1 + 19.62 $$
- Rearranging the equation
Rearranging the equation gives: $$ 0 = 30 t_1 - 2 \times 9.81 t_1 + 19.62 - 60 $$
Solving for $t_1$ simplifies it to: $$ 0 = (30 - 19.62) t_1 - 40.38 $$ $$ t_1 = \frac{40.38}{10.38} $$
- Calculating the time
Finally, compute $t_1$: $$ t_1 \approx 3.89 , \text{s} $$
The time that the second ball has been falling is: $$ t_2 = t_1 - 2 \approx 3.89 - 2 = 1.89 , \text{s} $$
The second ball will take approximately $1.89 , \text{s}$ to overtake the first ball.
More Information
When calculating the falling motion of two objects, it's important to consider their initial conditions and the total time of flight. The concepts of free fall and initial velocity are key in solving such kinematics problems.
Tips
- Ignoring Time Delay: Not considering the 2-second delay when throwing the second ball can lead to incorrect time calculations.
- Incorrectly Applying Gravity: Make sure to consistently use the acceleration due to gravity as $9.81 , \text{m/s}^2$.
- Neglecting to Set Distances Equal: Always remember that you need to equate the distances traveled by both balls at the moment they meet.
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