If the vertical initial speed of the ball is 2.5 m/sm/s as the cannon moves horizontally at a speed of 0.55 m/sm/s , how far from the launch point does the ball fall back into the cannon?

0.281 m

Explanation:

From vᵧ = uᵧ - gt

where vᵧ = final vertical component of the velocity

uᵧ = vertical component of the initial velocity = 2.5 m/s

g = 9.8 m/s²

At maximum height, vᵧ = 0 m/s

So,

Time to reach maximum height, t = uᵧ/g = 2.5/9.8 = 0.255 s

Total time of flight, T = 2 × time to reach maximum height

T = 2 × t = 2 × 0.255 = 0.510 s

Range of the cannon = uₓ T = 0.55 × 0.510 = 0.281 m

Note uₓ = horizontal component of the initial velocity.

b

explanation:

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c) both a and b.