A ball is dropped from rest from a height of 20.0m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?
Use the equations:
S=vi*t + (1/2)gt^2
For 1st ball:
20=0+(1/2)gt^2 => t=sqrt(40/g)
For 2nd ball:
20 = vi*(t-1) + (1/2)g(t-1)^2
vi = [20-(1/2)g(t-1)^2]/(t-1)
=(20-(9.81/2)*(t-1)^2)/(t-1)
=14.62 m/s downwards
To solve this problem, we can use the kinematic equations of motion. Let's break it down step by step:
Step 1: Find the time it takes for the first ball to reach the ground.
We can use the kinematic equation for displacement to find the time it takes for the first ball to fall:
h = (1/2) * g * t^2
Where h is the initial height (20m), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time. Solving for t:
20 = (1/2) * (-9.8) * t^2
t^2 = 20 / 4.9
t = √(20 / 4.9) = 2.02 seconds
So it will take 2.02 seconds for the first ball to reach the ground.
Step 2: Find the time it takes for the second ball to reach the ground.
Since the second ball is thrown vertically downwards, we can directly calculate the time it takes for it to reach the ground. The time would be 1 second less than the time it took for the first ball to reach the ground:
t2 = t1 - 1
t2 = 2.02 - 1
t2 = 1.02 seconds
So the second ball takes 1.02 seconds to reach the ground.
Step 3: Find the initial velocity of the second ball.
Now we can use the equation of motion for vertical displacement with constant acceleration:
h = v0 * t + (1/2) * g * t^2
Considering the final displacement is -20.0m (as it is moving downward), the initial displacement is 0, the acceleration is -9.8 m/s^2, and the time is 1.02 seconds, we can rearrange the equation to solve for the initial velocity of the second ball:
0 = v0 * 1.02 + (1/2) * (-9.8) * (1.02)^2
0 = v0 * 1.02 - 5 * (1.02)^2
5 * (1.02)^2 = v0 * 1.02
v0 = (5 * (1.02)^2) / 1.02
v0 = 5 * 1.02 = 5.1 m/s
Therefore, the initial velocity of the second ball must have been 5.1 m/s (downward).