A cannon shell is fired up in the air at an initial speed of 225m/s.
After how much timeis the shell at a height of 6.20 X 102 and heading down. What equation would be used to solve this?
Vf^2 = Vo^2 + 2(-9.8)d = 0,
(225)^2 - 19.6d = 0,
50625 - 19.6d = 0,
-19.6d = -50625,
d(up) = 2583m, max.
d = Vot + 0.5gt^2 = 2583 - 620 = 1963m
225t + 4.9t^2 = 1963,
4.9t^2 + 225t -1963 = 0,
Solve using Quadratic Formula and get:
t = 7.49958s, and t = - 48.526.
Use positive value of t.
t(down) = 7.49958s = time to go from
2583m to 620m.
To solve this problem, we can use the equations of motion for vertically launched projectiles. In this case, the shell is being launched up into the air and then it comes back down.
The equation that relates height, initial velocity, time, and acceleration for vertically launched projectiles is:
h = (v₀t) + (0.5gt²)
Where:
h - height of the object (6.20 × 10² meters in this case)
v₀ - initial velocity (225 m/s in this case)
t - time (which we are trying to find)
g - acceleration due to gravity (approximated as 9.8 m/s²)
Now, we can rearrange the equation to solve for time (t):
h = (v₀t) + (0.5gt²)
6.20 × 10² = (225t) + (0.5 × 9.8 × t²)
Since the shell is at the highest point when it starts coming down, its vertical velocity will be zero. This means we only need to consider the second term on the right side:
0.5 × 9.8 × t² = 6.20 × 10²
Now, we can solve for time (t):
4.9t² = 620
t² = 620 / 4.9
t² = 126.53
Taking the square root of both sides, we get:
t = √126.53
t ≈ 11.25 seconds
Therefore, the shell will be at a height of 6.20 × 10² meters and heading down after approximately 11.25 seconds.