A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 N on the ball. A) With what speed does the projectile leave the barrel of the cannon?
B) At what point does the ball have max speed?
C) What is the maximum speed?

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

falling water hydro = water .. spectcular places these plants are

f=ma newton's second law 7x8=56n