A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.620 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.
To solve this problem, we can use the equations of motion for free-falling objects. The equation we'll use is:
d = (1/2)gt^2
where:
- d is the distance the boy falls (from the top of the cliff to the water below)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken for the entire trip
Given that the boy falls halfway down to the water in 0.620 s, we can find the total time taken.
1. First, let's find the distance fallen at the halfway point.
d_halfway = (1/2)gt_halfway^2
Given that t_halfway = 0.620 s, we can substitute this value:
d_halfway = (1/2)(9.8 m/s^2)(0.620 s)^2
2. Now, we need to find the total distance fallen, which is twice the halfway distance.
d_total = 2 * d_halfway
3. Next, we can rearrange the equation to solve for the total time taken.
d_total = (1/2)gt_total^2
Rearranging gives:
t_total = sqrt((2 * d_total) / g)
4. Substitute the values of d_total and g into the equation:
t_total = sqrt((2 * (2 * d_halfway)) / g)
5. Calculate the final value for t_total.
By following these steps, you should be able to find out how much time passes during the boy's entire trip from the top down to the water.
To find the total time it takes for the boy to reach the water, we need to consider two parts of his motion: the time it takes to fall halfway down the cliff and the time it takes to fall the remaining distance.
Let's start by finding the time it takes for the boy to fall halfway down the cliff. We can use the kinematic equation for vertical motion:
d = (1/2) * g * t^2
where d is the distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since the boy falls halfway down the cliff, we have:
d = (1/2) * h
where h is the height of the cliff.
Given that t = 0.620 s, we can rearrange the equation to solve for h:
h = 2 * d
Now we can calculate the height of the cliff:
h = 2 * (1/2) * g * t^2
= g * t^2
Next, we want to find the time it takes for the boy to fall the remaining distance. Since the boy starts from rest at the top, we can use another kinematic equation:
d = (1/2) * g * t^2
where d is the remaining distance and t is the time. We need to solve this equation for t:
t = √(2 * d / g)
Since the remaining distance is also h, we can substitute h into the equation and solve for t:
t = √(2 * h / g)
Finally, to find the total time, we add the time it takes to fall halfway down the cliff and the time it takes to fall the remaining distance:
total time = t (for falling halfway down) + t (for falling remaining distance)
d1 = 0.5g*t^2 = 4.9*0.62^2 = 3.04 m = 1/2 the distance.
d = 0.5g*t^2 = 2*3.04 = 6.08 m.
4.9*t^2 = 6.08
t^2 = 1.24
t = 1.114 s. passed during entire trip.