A cannon fires a shell straight upward; 2.1 s after it is launched, the shell is moving upward with a speed of 20 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 4.7 s after the launch.
Well, let's see. A cannon firing a shell straight upward? That's quite an explosive way to express some velocity!
Now, to answer your question, we can use the good ol' kinematic equations. At launch, the speed of the shell would be the initial velocity. And since the shell has just been fired, we can assume it starts from rest. So, at launch, the speed would be 0 m/s. That's right, it's not going anywhere fast...yet!
Now, 2.1 seconds after launch, the shell is moving upward with a speed of 20 m/s. This means it has gained some velocity along the way. To calculate the magnitude of velocity at 4.7 seconds, we can use the equation:
v = u + at
where
v = final velocity,
u = initial velocity (at launch),
a = acceleration (due to gravity, which is approximately -9.8 m/s^2),
t = time.
Since we know the initial velocity is 0 m/s, the equation becomes:
v = 0 + (-9.8 × 2.1)
Solving that equation, we find that the velocity at 2.1 seconds is approximately -20.58 m/s. Negative because the shell is moving upward.
Now, to find the velocity at 4.7 seconds, we use the same equation:
v = u + at
This time, we know the initial velocity is -20.58 m/s (upward). So the equation becomes:
v = -20.58 + (-9.8 × 4.7)
Solving for v, we get approximately -64.14 m/s. Again, negative because the shell is still moving upward.
So, the magnitude of velocity at launch is 0 m/s, and at 4.7 seconds after launch, it's about 64.14 m/s (upward). That's one speedy shell! Just don't let it go to its head...or...err...fuse!
To find the speed of the shell at launch, we can use the equation for vertical velocity:
v = u + gt
where:
v = final velocity (20 m/s)
u = initial velocity (at launch)
g = acceleration due to gravity (-9.8 m/s²)
t = time (2.1 s)
Rearranging the equation, we have:
u = v - gt
Substituting the given values, we get:
u = 20 m/s - (-9.8 m/s²)(2.1 s)
u = 20 m/s + 20.58 m/s
u ≈ 40.58 m/s
Therefore, the speed (magnitude of velocity) of the shell at launch is approximately 40.58 m/s.
To find the speed of the shell 4.7 seconds after launch, we can use the same equation:
v = u + gt
where:
v = final velocity
u = initial velocity (40.58 m/s, obtained previously)
g = acceleration due to gravity (-9.8 m/s²)
t = time (4.7 s)
Substituting the given values, we get:
v = 40.58 m/s + (-9.8 m/s²)(4.7 s)
v ≈ 40.58 m/s - 45.86 m/s
v ≈ -5.28 m/s
Therefore, the speed (magnitude of velocity) of the shell 4.7 seconds after launch is approximately 5.28 m/s downward.
To find the speed (magnitude of velocity) of the shell at launch and 4.7 seconds after launch, we can use the equations of motion.
Let's break down the problem step-by-step:
Step 1: Determine the initial velocity of the shell at launch.
At launch, the shell is moving straight upward, and we know that the shell reached a speed of 20 m/s 2.1 seconds after launch. Since air resistance is negligible, the acceleration is constant (acceleration due to gravity, g ≈ 9.8 m/s²).
Using the equation of motion: v = u + at, where,
v = final velocity (20 m/s),
u = initial velocity (we need to find this),
a = acceleration (-9.8 m/s²), and
t = time (2.1 s).
Plugging in the known values, we can solve for the initial velocity (u):
20 m/s = u + (-9.8 m/s²)(2.1 s)
Simplifying the equation:
20 m/s = u - 20.58 m/s²
Solving for u, we get:
u = 20 m/s + 20.58 m/s²
u ≈ 40.58 m/s
Therefore, the initial velocity of the shell at launch is approximately 40.58 m/s.
Step 2: Determine the velocity of the shell 4.7 seconds after launch.
To find the velocity of the shell after 4.7 seconds, we will use the same equation of motion: v = u + at.
Using the same values for acceleration (a) and initial velocity (u), and substituting t = 4.7 s, we can solve for the final velocity (v):
v = 40.58 m/s + (-9.8 m/s²)(4.7 s)
v ≈ 40.58 m/s - 46.06 m/s
v ≈ -5.48 m/s
Therefore, the speed (magnitude of velocity) of the shell 4.7 seconds after launch is approximately 5.48 m/s. Note that the negative sign indicates that the shell is moving downward.
v = Vi - 9.81 t
20 = Vi - 9.81 (2.1)
solve for Vi
then
Use that Vi to get v at t = 4.7