An object is thrown downwards with a 2.5 m/s speed from a 9.6m height. Determine the initial velocity (in m/s) with which another object needs to be thrown upwards so that both hit each other at a 3.8m height from the ground.
Well, that sounds like a rather precarious game of catch! Let's see if we can figure out the initial velocity required for the other object to meet our downward-falling friend.
To solve this, we can use the equations of motion. Since we are dealing with vertical motion, we'll focus on the vertical components.
Let's call the initial velocity of the second object V2 and the time it takes for both objects to meet t.
For the first object (thrown downwards), we have:
Initial velocity (V1) = -2.5 m/s (negative because it's going downwards)
Final velocity (V1') = ? (since we're not given the final velocity explicitly)
Acceleration (a) = 9.8 m/s^2 (acceleration due to gravity)
Height (h) = 9.6 m
Using the equation of motion: V1' = V1 + at
Since V1' is the final velocity and it would be zero when it hits the 3.8 m height, we can rewrite the equation as:
0 = -2.5 + 9.8t
Simplifying, we get:
t ≈ 0.256 seconds
Now, let's move on to the second object (thrown upwards). We want it to reach a height of 3.8 m, so the equation for its vertical motion will be:
Height (h) = 3.8 m
Initial velocity (V2) = ?
Acceleration (a) = -9.8 m/s^2 (negative due to gravity)
Using the equation of motion: h = V2t + 0.5at^2
Plugging in the values, we get:
3.8 = V2(0.256) + 0.5(-9.8)(0.256)^2
3.8 = V2(0.256) - 0.318
Simplifying, we find:
V2 ≈ 15.763 m/s
So, the initial velocity of the second object should be approximately 15.763 m/s upwards to meet the first object at a height of 3.8 m. Just make sure to warn your friends about the daring game of catch you have planned!
To solve this problem, we can use the equations of motion. The equations we will need are:
1. Displacement (d) = initial velocity (u) * time (t) + 0.5 * acceleration (a) * time (t) squared
2. Final velocity (v) = initial velocity (u) + acceleration (a) * time (t)
3. Final velocity squared (v^2) = initial velocity squared (u^2) + 2 * acceleration (a) * displacement (d)
Given:
Initial velocity of the first object (u1) = -2.5 m/s (since it is thrown downwards)
Height of the first object (h1) = 9.6 m
Final height (h2) = 3.8 m
We need to find the initial velocity of the second object (u2).
Let's begin by finding the time it takes for the first object to hit the ground. We can use equation 1 with h1 = -9.6 m (since it is downwards) and d = -9.6 m:
-9.6 = -2.5t + 0.5 * 9.8 * t^2
Simplifying the equation, we get:
4.9t^2 - 2.5t - 9.6 = 0
Solving this quadratic equation, we find two values for t: t1 ≈ 1.778 s and t2 ≈ 2.176 s. Since the object is thrown downwards, we need the smaller time value, so t1 ≈ 1.778 s.
Now, we can use equation 2 to find the final velocity (v1) of the first object just before hitting the ground:
v1 = u1 + a * t1
= -2.5 + 9.8 * 1.778
≈ 15.81 m/s
Next, we need to find the time it takes for the second object to reach the final height of 3.8 m. We can use equation 1 with h1 = 0 m (since it starts from the ground) and d = 3.8 m:
3.8 = u2 * t + 0.5 * (-9.8) * t^2
Simplifying the equation, we get:
-4.9t^2 + u2t + 3.8 = 0
To find u2, we need to solve this equation. However, we already have the value of v1 (which will be u2 for the second object), so we can substitute it into the equation:
-4.9t^2 + 15.81t + 3.8 = 0
Solving this quadratic equation, we find two values for t: t1 ≈ 0.542 s and t2 ≈ 1.489 s. Since the object is thrown upwards, we need the larger time value, so t2 ≈ 1.489 s.
Now that we have t2, we can find the initial velocity of the second object (u2):
u2 = v1 - a * t2
= 15.81 - 9.8 * 1.489
≈ -0.83 m/s
Therefore, the initial velocity with which the second object needs to be thrown upwards is approximately -0.83 m/s.