Question

What is the acceleration of Timmy from the previous problem?

A. 0.540 m/s²
B. 0.640 m/s²
C. 0.740 m/s²
D. 0.840 m/s²

A gauss cannon will accelerate a projectile at an incoming asteroid at 22.3 m/s². If the projectile begins from rest and the length of the cannon is 114 m, what is the final velocity of the projectile as it leaves the cannon?

A. 51.3 m/s
B. 71.3 m/s
C. 3,043 m/s
D. 5,084 m/s

Answer 1

The final velocity of the projectile as it leaves the cannon is approximately 71.8 m/s.

Step-by-step explanation:

To find the final velocity of the projectile as it leaves the cannon, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is 0 m/s since the projectile starts from rest), a is the acceleration (which is 22.3 m/s^2), and s is the distance (which is 114 m).

First, we rearrange the equation to solve for v:

v^2 = 2as

Then, substitute the given values:

v^2 = 2(22.3 m/s^2)(114 m)

v^2 = 5146.8 m^2/s^2

Taking the square root of both sides, we find:

v = sqrt(5146.8 m^2/s^2) ≈ 71.8 m/s

So, the final velocity of the projectile as it leaves the cannon is approximately 71.8 m/s.