4) a cannon shoots a cannonball of mass 5kg vertically upward from the mouth of a cannon with muzzle velocity of 7 m/s. at a height of 25 m above the mouth of the cannon, the cannonball attaches to another mass (1kg) and they continue to travel up. what is the maximum height that both masses will reach?

h =220 m

Explanation:

Given that

u = 7 m/s

Even mass will attach but this will not produce any effect on the maximum height of the ball.Because in energy conservation the effect of mass does not present.

So the final speed of the ball will be zero at the maximum height.

v² = u² - 2 g (25 + h)

0 = 7² - 2 x 10 (25 +h)

49 = 20 ( 25 +h)

49 = 500 +20 h

Here h comes out negative that is why we are taking the 70 m/s in place of 7 m/s.

0 = 70² - 2 x 10 (25 +h)         ( take g =10 m/s²)

4900 = 20 ( 25 +h)

4900 = 500 +20 h

4900- 500 = 20 h

4400 = 20 h

440 = 2 h

h =220 m

answer; how far out of place which the object finished relative to the starting location;