A cannon fires a shell straight upward; 2.4 s after it is launched, the shell is moving upward with a speed of 18 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.4 s after the launch.
What is the velocity at launch?
Velocity after 5.4s?
vf=vi-gt
18=vi-9.8(2.4) solve for vi
using that vi, solve at time 5.4sec
To find the velocity at launch, we can use the concept of projectile motion. When the cannon fires the shell straight upward, the only force acting on it is gravity, which causes it to accelerate downward at a constant rate.
The equation we can use to find the velocity at any given time is:
v = u + gt
Where:
v = final velocity
u = initial velocity (velocity at launch)
g = acceleration due to gravity (approximately -9.8 m/s^2)
t = time
In this case, we know that the shell is moving upward with a speed of 18 m/s at 2.4 seconds after launch. However, at the highest point of its trajectory, the shell momentarily stops before starting to fall back down. So, we can assume that the final velocity at this point is zero.
Plugging the given values into the equation, we have:
0 = u + (-9.8 m/s^2) * 2.4 s
Simplifying the equation:
0 = u - 23.52 m/s
Rearranging the equation to solve for u:
u = 23.52 m/s
Therefore, the velocity at launch is 23.52 m/s.
Now, let's find the velocity 5.4 seconds after the launch.
Using the same equation:
v = u + gt
Given values:
u = 23.52 m/s (the velocity at launch)
g = -9.8 m/s^2 (acceleration due to gravity)
t = 5.4 s (time after launch)
Plugging these values into the equation:
v = 23.52 m/s + (-9.8 m/s^2) * 5.4 s
Simplifying the equation:
v = 23.52 m/s - 52.92 m/s
v = -29.4 m/s
The negative sign indicates that the shell is moving downward, as it would during the descending part of its trajectory.
Therefore, the velocity 5.4 seconds after the launch is -29.4 m/s.