Answer: 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.
