Two stones are thrown simultaneously. One is thrown straight upward from the base of a cliff and the other is thrown straight downward from the top of the cliff. The height of the cliff is 6 m. The stones are thrown with the same speed of 9.49 m/s. Find the location above the base of the cliff where the stones cross paths.
Just 6 meters high, oh well
downward problem
z = 6 - 9.49 t - 4.9 t^2
upward problem
z = 0 + 9.49 t - 4.9 t^2
so at collision
6 - 9.49 t = 9.49 t
6 = 19 t
t = 0.316 seconds
z = 9.49 (0.316) -4.9 (0.316)^2
= 3 - .5 = 2.5 meters
wouldnt the downward problem have +4.9 t^2 because you distribute the minus sign?
Nope, not accelerating upwards
but the upward ball is so i was thinking we would do
downwards distance = 6- (upwards)
= 6 - (9.49t -4.9t^2)
= 6 - 9.49 t + 4.9t^2
V = Vi + a t
h = Hi + Vi t + (1/2) a t^2
here starting from the ground up
Vi is up
a is DOWN (-9.81), in other words it slows down as it rises, and after a while stops rising.
gravity speeds it up coming down, and slows it going up
To find the location above the base of the cliff where the stones cross paths, we first need to determine the time it takes each stone to reach that point.
Let's start by examining the stone thrown upward. To do this, we'll use the equation for the displacement of an object in free fall:
h = v0t - (1/2)gt^2
Where:
h = height of the cliff (6 m)
v0 = initial velocity of the stone thrown upward (9.49 m/s)
t = time taken to reach the point of intersection
g = acceleration due to gravity (9.8 m/s^2)
Since the stone is thrown upward, its initial velocity is positive (+9.49 m/s).
Setting the equation equal to 0 (since the stone reaches the point of intersection at its maximum height), we have:
0 = v0t - (1/2)gt^2
Simplifying the equation, we get:
(1/2)gt^2 = v0t
Dividing both sides of the equation by t, we obtain:
(1/2)gt = v0
Now, let's consider the stone thrown downward. Using the same equation, we have:
h = v0t + (1/2)gt^2
Again, setting the equation equal to 0 (since the stone reaches the point of intersection at the base of the cliff), we get:
0 = v0t + (1/2)gt^2
Simplifying, we have:
(1/2)gt^2 = -v0t
Dividing by t, we have:
(1/2)gt = -v0
Since both stones reach the point of intersection at the same time, we can equate the two expressions for (1/2)gt:
v0 = -v0
Solving for t, we find:
t = 2v0/g
Substituting the values given:
v0 = 9.49 m/s
g = 9.8 m/s^2
t = 2(9.49 m/s) / 9.8 m/s^2 ≈ 1.93 s
Therefore, both stones reach the point of intersection approximately 1.93 seconds after being thrown.
To find the location above the base of the cliff where the stones cross paths, we need to calculate the distance traveled by each stone in that time.
Using the equation for displacement, s = v0t - (1/2)gt^2, we find:
For the stone thrown upward:
s1 = v0t - (1/2)gt^2
s1 = 9.49 m/s * 1.93 s - (1/2) * 9.8 m/s^2 * (1.93 s)^2
s1 ≈ 9.13 m
For the stone thrown downward:
s2 = v0t + (1/2)gt^2
s2 = -9.49 m/s * 1.93 s + (1/2) * 9.8 m/s^2 * (1.93 s)^2
s2 ≈ -17.41 m
Since we're interested in the distance above the base of the cliff, we can add the height of the cliff (6 m) to the positive value of s1:
Distance above the base = s1 + height of cliff
Distance above the base = 9.13 m + 6 m
Distance above the base ≈ 15.13 m
Therefore, the stones cross paths approximately 15.13 meters above the base of the cliff.