Awheeled car with a spring-loaded cannon fires a metal ball vertically. if the vertical initial speed of the ball is 5 m/s as the cannon moves horizontally at a speed of .75 m/s. how far from the launch point does the ball fall back into the cannon? what would happen if the cannon were accelerating

The important physics fact to take in account in this problem is that the horizontal velocity of the ball is supposed to remain constant and is the same horizontal velocity of the cannon: 0.75 m/s (valid for the first part of the problem).

Then what you need to calculate is the time of flight (since the ball is fired up until it comes back to the level of the cannon).

There are several formulas that let you calculate this.

I will use = t of flight = 2*time until the highest point = 2Vo / g , where Vo is the initial vertical velocity.

t of flight = 2* 5 m/s / 9.8 m/s^2 = 1.02 s.

Now calculate the horizontal distance runned when t = 1.02 s and v = 0.75  m/s

d = vt = 0.75 m/s * 1.02 s = 0.77m

That is the answer of the first part.

Second part:

If the cannon were accelerating the ball would miss it when came back to the ground. Then, the ball would hit the ground, and the cannon would have left behind the point of impact.

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