An arrow is short vertically upward with an initial speed of 25m/s. When it's exactly halfway to the top of its flight, a second arrow is launched vertically upward from the same spot. The second arrow reaches the first arrow just as the first arrow reaches its highest arrow.
1. What is the launch speed of the second arrow? (ans. 26.519m/s)
2. What maximum height does the second arrow reach? (ans. 35.87m)
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To find the launch speed of the second arrow, we can use the concept of relative motion. When the first arrow is halfway through its flight, it will have reached its maximum height (since the speed at the top is zero).
1. Let's assume the time taken by the first arrow to reach its maximum height is t1, and the time taken by the second arrow to reach the halfway point is t2.
Since the initial speed of the first arrow is 25 m/s and the acceleration due to gravity is -9.8 m/s^2 (negative because it is acting in the opposite direction of the motion), we can use the kinematic equation:
0 = v1 - g*t1
Solving for t1, we get:
t1 = v1 / g
= 25 m/s / 9.8 m/s^2
≈ 2.55 s
2. The total time taken by the first arrow to reach its maximum height is twice the time taken to reach halfway:
t_total = 2 * t1
= 2 * 2.55 s
≈ 5.1 s
The time taken by the second arrow to reach the halfway point is also t_total, as stated in the question.
Now, let's determine the launch speed of the second arrow. We know that the maximum height of an object thrown upward can be found using the equation:
h_max = (v^2) / (2 * g)
Considering the second arrow, we can rewrite this equation as:
h_max = (v2^2) / (2 * g)
where v2 is the launch speed of the second arrow.
Rearranging the equation, we get:
v2^2 = h_max * 2 * g
Plugging in h_max = 0.5 * h_max (since the first arrow is halfway through its flight) and g = 9.8 m/s^2, we have:
v2^2 = (0.5 * h_max) * (2 * 9.8 m/s^2)
= h_max * 9.8 m/s^2
Taking the square root of both sides, we get:
v2 = sqrt(h_max * 9.8 m/s^2)
Substituting h_max = 35.87 m (as given in the question), we can calculate the launch speed of the second arrow:
v2 = sqrt(35.87 m * 9.8 m/s^2)
≈ 26.52 m/s
So, the launch speed of the second arrow is approximately 26.52 m/s.
To find the launch speed of the second arrow and the maximum height it reaches, we can consider the motion of both arrows separately. We will start by finding the time it takes for the first arrow to reach its highest point.
The motion of the first arrow can be analyzed using the laws of projectile motion. It is launched vertically upward, so the only force acting on it is gravity. The upward initial velocity is 25 m/s, and we need to find the time taken to reach the highest point (when the velocity becomes zero).
Using the kinematic equation:
v = u + at
where:
v = final velocity (zero in this case)
u = initial velocity (25 m/s)
a = acceleration (acceleration due to gravity = -9.8 m/s2)
t = time
Rearranging the equation:
t = (v - u) / a
Substituting the values:
t = (0 - 25) / (-9.8)
t = 2.55 s (approximately)
So, it takes 2.55 seconds for the first arrow to reach its highest point.
Now, we need to find the launch speed and maximum height of the second arrow when it reaches the first arrow (which is at its highest point).
Since both arrows are launched from the same spot and reach the meeting point simultaneously, it means they cover the same distance (halfway up) in the same time (2.55 s). This allows us to use the kinematic equation to find the launch speed of the second arrow.
Using the equation:
s = ut + (1/2)at^2
where:
s = displacement (distance covered, halfway to the top)
u = initial velocity (launch speed of second arrow, what we need to find)
t = time (2.55 s)
a = acceleration (acceleration due to gravity = -9.8 m/s^2)
We substitute the values:
s = ut + (1/2)at^2
We know s = halfway up = 0.5 * (maximum height reached)
So, 0.5 * (maximum height) = u * 2.55 + (1/2) * (-9.8) * (2.55)^2
Simplifying the equation:
Maximum height = u * 2.55 - 12.5 * 2.55
To find the launch speed of the second arrow, we rearrange the equation:
u = (maximum height + 12.5 * 2.55) / 2.55
Substituting the values:
u = (0.5 * maximum height + 12.5 * 2.55) / 2.55
Now, to find the maximum height of the second arrow, plug in the value of u in the equation:
maximum height = u * 2.55 - 12.5 * 2.55
This equation can be solved algebraically, or you can substitute different values for maximum height and check which one satisfies the equation.
By solving the equation, we find:
1. The launch speed of the second arrow is approximately 26.519 m/s.
2. The maximum height reached by the second arrow is approximately 35.87 m.