A cannon shell is fired straight up from the ground at an initial speed of 225 m/s. After how much time is the shell at a height of 6.20 x 102 m above the ground and moving downward? (a) 2.96 s (b) 17.3 s (c) 25.4 s (d) 33.6 s (e) 43.0 s
5
620 = -4.9 t^2 + 225 t
0 = -4.9 t^2 + 225 t - 620
solve with quadratic formula
... you want the largest solution (on the way down)
To find out how much time the cannon shell takes to reach a certain height and moving downward, we can use the equation of motion:
h = v₀t + (1/2)gt²
where:
h = displacement (height above the ground)
v₀ = initial velocity (225 m/s)
t = time taken
g = acceleration due to gravity (-9.8 m/s², assuming upward as positive)
Since we are trying to find the time taken when the shell is at a height of 6.20 x 10² m, we can substitute the given values into the equation and solve for t:
6.20 x 10² = (225)t + (1/2)(-9.8)t²
Rearranging the equation:
0 = (1/2)(-9.8)t² + 225t - 6.20 x 10²
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
where:
a = 1/2(-9.8)
b = 225
c = -6.20 x 10²
Plugging in the values into the quadratic formula:
t = (-225 ± √(225² - 4(1/2)(-9.8)(-6.20 x 10²))) / (2(1/2)(-9.8))
Simplifying the equation:
t = (-225 ± √(225² + 1.2 x 10²)) / (-9.8)
Calculating the values inside the square root:
t = (-225 ± √(50625 + 120)) / -9.8
t = (-225 ± √50745) / -9.8
Using a calculator to find the square root:
t ≈ (-225 ± 225.17) / -9.8
Simplifying further:
t ≈ (-450.17) / (-9.8) OR t ≈ 0.17 / (-9.8)
Calculating the answers:
t ≈ 45.92 s OR t ≈ -0.02 s
Since the time cannot be negative, we discard the negative value. Therefore, the time taken for the shell to reach a height of 6.20 x 10² m and moving downward is approximately 45.92 seconds.
None of the answer choices provided match this value exactly, so it seems that none of the given options are correct.
To find the time when the cannon shell is at a height of 6.20 x 10^2 m above the ground and moving downward, we can use the equations of motion.
First, let's find the time it takes for the cannon shell to reach its highest point. To do this, we can use the equation:
v = u + at
Where:
v = final velocity (we know the velocity at its highest point is 0 m/s)
u = initial velocity (225 m/s)
a = acceleration (due to gravity, it is -9.8 m/s^2, negative because it acts opposite to the direction of motion)
t = time
0 = 225 - 9.8t
Solving this equation, we get:
t = 225 / 9.8
t ≈ 22.96 s
So, it takes approximately 22.96 seconds for the cannon shell to reach its highest point.
Next, let's find the total time it takes for the cannon shell to reach a height of 6.20 x 10^2 m above the ground and moving downward.
We can use another equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement (6.20 x 10^2 m)
u = initial velocity (225 m/s)
a = acceleration (due to gravity, it is -9.8 m/s^2, negative because it acts opposite to the direction of motion)
t = time
6.20 x 10^2 = 225t - (1/2)(9.8)t^2
Rearranging the equation and converting it to a quadratic equation:
(1/2)(9.8)t^2 - 225t + 6.20 x 10^2 = 0
Solving this equation for t will give us the total time.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 1/2(9.8) = 4.9
b = -225
c = 6.20 x 10^2
t = (-(-225) ± √((-225)^2 - 4(4.9)(6.20 x 10^2))) / (2(4.9))
After evaluating this equation, we find two values of t:
t ≈ 25.4 s and t ≈ -2.96 s
Since the cannon shell is moving downward, we discard the negative time value.
Therefore, the cannon shell is at a height of 6.20 x 10^2 m above the ground and moving downward after approximately 25.4 seconds.
The correct answer is (c) 25.4 s.