A ball is thrown vertically upward with a velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. at what position above the ground will they meet?
The time of the first ball motion to the top point is
v=v(o1) –gt, v=0,
t=v(o1) /g=18/9.8 = 1.84 s.
The maximum height of the 1st ball is
H= v(o1) •t –gt²/2=16.53 m.
When the 2nd ball starts, the 1st ball is on its return trip during 2-1.84 =0.16 s.
It covers the distance hₒ=g(0.16)²/2 =0.125 m. and has the downward velocity
v(1)=g(0.16)=1.568 m/s.
At this instant the distance between two balls is 16.53-0.125=16.4 m, and two balls are moving towards each other: the first ball accelerates downward with initial velocity v(1)=1,568 m/s, the second ball decelerates upward at initial velocity v(2)=13.5 m/s. They cover the distances:
the1st ball h1= v1•t+gt²/2; the 2nd ball h2=v2•t-gt²/2.
h1+h2= 16.4 m.
=> v1•t+gt²/2+ v2•t-gt²/2 =16.4
t=16.4/(v1+v2)=16.4/(1.568+13.5)=1.088 s.
The point where the balls mett is located
h= v2•t-gt²/2 =13.5•1.088-g•(1.088)²/2 =8.89 m (above the ground)
To find the position where the two balls meet, we need to determine the time it takes for each ball to reach that position.
Let's start by figuring out the time it takes for the first ball to reach the meeting point. We know that the initial velocity of the first ball is 18 m/s, and it is thrown vertically upwards, which means it will experience negative acceleration due to gravity.
The equation we can use to determine the time it takes for the ball to reach a certain height is given by:
y = ut + (1/2) * a * t^2
Where:
- y is the displacement or height (in this case, the position where the balls meet)
- u is the initial velocity
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
For the first ball:
y1 = 0 (since it is thrown from the ground)
u1 = 18 m/s
a = -9.8 m/s^2
Since we're looking for the time, we can rearrange the equation:
0 = (18) * t1 + (1/2) * (-9.8) * t1^2
Simplifying the equation, we get:
-4.9t1^2 + 18t1 = 0
Now, let's solve for t1:
t1 * (-4.9t1 + 18) = 0
Either t1 = 0 or -4.9t1 + 18 = 0
Ignoring the t1 = 0 case (which corresponds to the initial throwing time), solve for t1:
-4.9t1 + 18 = 0
-4.9t1 = -18
t1 = -18 / -4.9
t1 ≈ 3.67 seconds
So, it takes approximately 3.67 seconds for the first ball to reach the meeting point.
Now, let's find the time it takes for the second ball to reach the meeting point. The initial velocity of the second ball is 13.5 m/s.
Using the same equation, we have:
y2 = 0 (since it is thrown from the ground)
u2 = 13.5 m/s
a = -9.8 m/s^2
0 = (13.5) * t2 + (1/2) * (-9.8) * t2^2
Simplifying and solving for t2:
-4.9t2^2 + 13.5t2 = 0
t2 * (-4.9t2 + 13.5) = 0
Either t2 = 0 or -4.9t2 + 13.5 = 0
Ignoring the t2 = 0 case, solve for t2:
-4.9t2 + 13.5 = 0
-4.9t2 = -13.5
t2 = -13.5 / -4.9
t2 ≈ 2.76 seconds
So, it takes approximately 2.76 seconds for the second ball to reach the meeting point.
Now that we have the times for each ball, let's calculate the position where they meet:
y = ut + (1/2) * a * t^2
For the first ball:
y1 = (18) * (3.67) + (1/2) * (-9.8) * (3.67)^2
For the second ball:
y2 = (13.5) * (2.76) + (1/2) * (-9.8) * (2.76)^2
Evaluating these equations will give us the position above the ground where the two balls meet.