1. To determine the vertical position of a ball thrown upward, we can use the equation:
y(t) = y0 + v0t - (1/2)gt^2
where:
y(t) = vertical position of the ball at time t
y0 = initial vertical position (in this case, 100 m)
v0 = initial velocity (in this case, 30.1 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
For every interval of 1 s until 10.0 seconds, we can plug in the value of t and solve for y(t):
t = 1s: y(1) = 100 + 30.1(1) - (1/2)(9.8)(1^2) = 129.3 m
t = 2s: y(2) = 100 + 30.1(2) - (1/2)(9.8)(2^2) = 152.9 m
t = 3s: y(3) = 100 + 30.1(3) - (1/2)(9.8)(3^2) = 170.7 m
t = 4s: y(4) = 100 + 30.1(4) - (1/2)(9.8)(4^2) = 183.6 m
t = 5s: y(5) = 100 + 30.1(5) - (1/2)(9.8)(5^2) = 191.5 m
t = 6s: y(6) = 100 + 30.1(6) - (1/2)(9.8)(6^2) = 195.4 m
t = 7s: y(7) = 100 + 30.1(7) - (1/2)(9.8)(7^2) = 195.3 m
t = 8s: y(8) = 100 + 30.1(8) - (1/2)(9.8)(8^2) = 191.2 m
t = 9s: y(9) = 100 + 30.1(9) - (1/2)(9.8)(9^2) = 183.1 m
t = 10s: y(10) = 100 + 30.1(10) - (1/2)(9.8)(10^2) = 170 m
2. a. To find the time it takes for the container to hit the ground, we can use the equation:
y(t) = vertical position of the container at time t (0 m)
y0 = initial vertical position (50 m)
v0 = initial velocity (0 m/s, since it is dropped)
Plugging in the values and setting y(t) = 0:
0 = 50 + (1/2)(9.8)t^2
t = √(50/(1/2)(9.8)) = √(50/4.9) = √(10.2) ≈ 3.2 s
b. To find the speed of the container as it hits the ground, we can use the equation:
v(t) = v0 + gt
v(t) = velocity of the container at time t
PA BRAINLIESTS PO
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Answers & Comments
1. To determine the vertical position of a ball thrown upward, we can use the equation:
y(t) = y0 + v0t - (1/2)gt^2
where:
y(t) = vertical position of the ball at time t
y0 = initial vertical position (in this case, 100 m)
v0 = initial velocity (in this case, 30.1 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
For every interval of 1 s until 10.0 seconds, we can plug in the value of t and solve for y(t):
t = 1s: y(1) = 100 + 30.1(1) - (1/2)(9.8)(1^2) = 129.3 m
t = 2s: y(2) = 100 + 30.1(2) - (1/2)(9.8)(2^2) = 152.9 m
t = 3s: y(3) = 100 + 30.1(3) - (1/2)(9.8)(3^2) = 170.7 m
t = 4s: y(4) = 100 + 30.1(4) - (1/2)(9.8)(4^2) = 183.6 m
t = 5s: y(5) = 100 + 30.1(5) - (1/2)(9.8)(5^2) = 191.5 m
t = 6s: y(6) = 100 + 30.1(6) - (1/2)(9.8)(6^2) = 195.4 m
t = 7s: y(7) = 100 + 30.1(7) - (1/2)(9.8)(7^2) = 195.3 m
t = 8s: y(8) = 100 + 30.1(8) - (1/2)(9.8)(8^2) = 191.2 m
t = 9s: y(9) = 100 + 30.1(9) - (1/2)(9.8)(9^2) = 183.1 m
t = 10s: y(10) = 100 + 30.1(10) - (1/2)(9.8)(10^2) = 170 m
2. a. To find the time it takes for the container to hit the ground, we can use the equation:
y(t) = y0 + v0t - (1/2)gt^2
where:
y(t) = vertical position of the container at time t (0 m)
y0 = initial vertical position (50 m)
v0 = initial velocity (0 m/s, since it is dropped)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Plugging in the values and setting y(t) = 0:
0 = 50 + (1/2)(9.8)t^2
t = √(50/(1/2)(9.8)) = √(50/4.9) = √(10.2) ≈ 3.2 s
b. To find the speed of the container as it hits the ground, we can use the equation:
v(t) = v0 + gt
where:
v(t) = velocity of the container at time t
PA BRAINLIESTS PO