You and two friends are on top of the library 30 m above the ground. At the same time that you drop a ball, from rest, one of your friends tosses a ball straight up, giving it an initial velocity of 15 m/s, and your other friend tosses a ball straight down, giving it an initial velocity of -15m/s. Note that each ball, once released from the tosser's hand will accelerate at -9.8m/s^2.
a) Assume the t=0 sec corresponds to when the three balls were released, what times will each ball strike the ground?
b) When the ball was tossed downward is 15 m above the ground, what will be the positions of the other two balls?
To solve this problem, we can use the equations of motion to calculate the time it takes for each ball to strike the ground and their positions at a specific time. Let's start with part (a).
a) To find the time it takes for each ball to strike the ground, we can use the equation for vertical displacement:
d = v₀t + (1/2)gt²
Where:
- d is the vertical displacement (in this case, it will be the height above the ground),
- v₀ is the initial velocity of the ball (0 m/s for the dropped ball, 15 m/s for the one tossed upward, and -15 m/s for the one tossed downward),
- g is the acceleration due to gravity (-9.8 m/s²), and
- t is the time.
Since we are looking for the time when the balls strike the ground, we can set d = 0 and solve for t.
1) Ball dropped: d = 0
0 = 0(t) + (1/2)(-9.8)(t)²
0 = (1/2)(-9.8)(t)²
-9.8(t)² = 0
t = 0 (No time is required for the dropped ball to strike the ground because it is already on the ground.)
2) Ball tossed upward: d = 30 m
30 = 15(t) + (1/2)(-9.8)(t)²
0 = (1/2)(-9.8)(t)² + 15(t) - 30
To solve the quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Using the values a = (1/2)(-9.8), b = 15, and c = -30, we can substitute them into the quadratic formula:
t = [ -15 ± √(15² - 4 * (1/2)(-9.8)(-30)) ] / [ 2 * (1/2)(-9.8) ]
Simplifying further:
t = [ -15 ± √(225 + 588) ] / ( -9.8 )
t ≈ [ -15 ± √(813) ] / ( -9.8 )
Using a calculator, we can find the two possible values for t. One value will be negative, which doesn't make sense in this context because you can't have a negative time. So, we only consider the positive value.
t ≈ 4.78 seconds (approx.)
3) Ball tossed downward: d = 15 m
15 = -15(t) + (1/2)(-9.8)(t)²
0 = (1/2)(-9.8)(t)² - 15(t) + 15
Again, we need to solve the quadratic equation using the quadratic formula:
t = [ 15 ± √(15² - 4 * (1/2)(-9.8)(15)) ] / [ 2 * (1/2)(-9.8) ]
Simplifying further:
t = [ 15 ± √(225 + 294) ] / ( -9.8 )
t ≈ [ 15 ± √(519) ] / ( -9.8 )
Once again, by using a calculator, we find the two possible values for t. Similarly, we only consider the positive value.
t ≈ 7.58 seconds (approx.)
So, to summarize the times when each ball strikes the ground:
- The ball dropped directly strikes the ground at t = 0 seconds.
- The ball tossed upward strikes the ground at t ≈ 4.78 seconds.
- The ball tossed downward strikes the ground at t ≈ 7.58 seconds.
b) Now, let's find the positions of the other two balls when the ball tossed downward is 15 m above the ground. This corresponds to the time t ≈ 7.58 seconds.
For the ball dropped, we know that it will already be on the ground since it fell instantly. So its position will be 0 m.
For the ball tossed upward, we can use the vertical displacement equation again:
d = v₀t + (1/2)gt²
Using v₀ = 15 m/s and t = 4.78 seconds:
d = 15(4.78) + (1/2)(-9.8)(4.78)²
d ≈ 71.7 meters (approx.)
Therefore, the ball tossed upward will be approximately 71.7 meters above the ground.
To summarize:
- The ball dropped will be on the ground (0 m above).
- The ball tossed upward will be approximately 71.7 meters above the ground.
- The ball tossed downward will be 15 m above the ground.