A cannon fires a shell straight upward; 1.7 s after it is launched, the shell is moving upward with a speed of 19 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.
v = Vi - 9.81 t
at 1.7 s
19 = Vi - 9.81 (1.7)
solve for Vi, launch speed
at 5.4 s
v = Vi - 9.81(5.4)
To solve this problem, we need to use the equations of motion for an object in free fall. The object is initially launched with an unknown speed, and then we need to find its speed at two different times (1.7 seconds and 5.4 seconds).
First, let's find the initial speed (or the speed at launch) of the shell. We'll use the equation:
v = u + gt
where
v is the final velocity (19 m/s),
u is the initial velocity (which we want to find),
g is the acceleration due to gravity (-9.8 m/s^2), and
t is the time (1.7 seconds).
Rearranging the equation, we have:
u = v - gt
Substituting the known values, we get:
u = 19 m/s - (-9.8 m/s^2) * 1.7 s
u = 19 m/s + 16.66 m/s
u = 35.66 m/s
Therefore, the speed (magnitude of velocity) of the shell at launch is 35.66 m/s.
Next, let's find the speed of the shell 5.4 seconds after the launch. During this time, the shell will be moving upward and then start falling back down. At its peak height, the shell's velocity will be momentarily zero.
To find the speed at 5.4 seconds, we'll use the equation:
v = u + gt
Using the known values:
u = 35.66 m/s (speed at launch),
g = -9.8 m/s^2 (acceleration due to gravity), and
t = 5.4 s (time since launch).
v = 35.66 m/s - 9.8 m/s^2 * 5.4 s
v = 35.66 m/s - 52.92 m/s
v = -17.26 m/s
Therefore, the speed (magnitude of velocity) of the shell 5.4 seconds after the launch is 17.26 m/s.
Note: The negative sign indicates that the shell is moving downward.