Answer:
Shown below;
Explanation:
From Newton's third law action and reaction.
Hence the force initiated by the spring is defined as;
F= K×E ; K- spring constant
E- distance moved by the spring.
F = 8.0 × 0.05= 0.4N; 0.05 m = 5cm.
The Canon would move out the gun with a speed of the net force.
The net force = 0.05 -0.032= 0.018N
This force = mass × velocity/ time
= mass × distance /time^2
0.018 =0.0053×0.15/t^2
t^2 = 0.0053×0.15/0.018=0.0442s
t=√0.0442=0.21
Velocity = distance/time
=0.15/0.21=0.71m/s
B. it sustains Maximum speed when it reaches there ground.
C. We assume the gun leaves at 45° to get the maximum distance which is the maximum height.
H = u^2sin2A/2g
= 0.71^2 sin 2×45/ 2×9.8
=0.5041/18.6=0.0271m
From Newton's law
V2 =u2+2gh
V2 = 0.71^2 + 2×9.8×0.0271
= 0.5041 + 0.53116
V=√1.03526=1.0175 = 1.02m/s
