A cannon fires a shell straight upward; 2.3 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) in meters per second of the shell at launch and 4.7 s after the launch.

The launch speed is 43m/s

4.7secs after launch speed is 4m/s

Explanation: To solve this we use the first equation of motion but in this case our acceleration would be -10m/s² since we are going upwards against gravity(launch).

Vf = Vi - a*t

Where Vf is the final velocity after launch, Vi is the initial velocity at launch, t is time in secs then a is acceleration

a. From the question

t = 2.3secs

Vf = 20m/s

a = -10m/s²

Substituting into the above equation we have that,

20= Vi - 10* 2.3

20 = Vi - 23

Vi = 20+23

Vi = 43 m/s

Which is the speed at launch.

b. The magnitude of speed (Vf) 4.7 sec after launch is calculated as follows using same procedure but here Vi is 43m/s as calculated

Vf = 43 - 10*4.7

Vf = 43 - 47

Vf = -4m/s

But since we are asked to find the magnitude we neglect the negative sign.

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~