A projectile of mass 0.539 kg is shot from a

cannon. The end of the cannon’s barrel is
at height 6.4 m, as shown in the figure. The
initial velocity of the projectile is 9 m/s .
How long does it take the projectile to hit the
ground?
Answer in units of s

To find the time it takes for the projectile to hit the ground, we can use the equations of motion.

1. First, we need to set up the problem. Since the projectile is fired horizontally, the initial vertical velocity is zero. The only force acting in the vertical direction is the force due to gravity.

2. We can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where h is the height of the projectile, u is the initial vertical velocity (which is zero in this case), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

3. Rearranging the equation, we get:
6.4 = 0 + (1/2) * 9.8 * t^2

4. Simplifying the equation:
6.4 = 4.9t^2

5. Dividing both sides of the equation by 4.9:
(6.4/4.9) = t^2

6. Taking the square root of both sides:
sqrt(6.4/4.9) = t

7. Evaluating the square root:
t ≈ sqrt(1.31) ≈ 1.14 seconds

So, it takes approximately 1.14 seconds for the projectile to hit the ground.

To find the time it takes for the projectile to hit the ground, we can use the kinematic equation:

y = y0 + v0y*t - (1/2) * g * t^2

Where:
- y is the final vertical position (0 m in this case, as it hits the ground)
- y0 is the initial vertical position (6.4 m in this case)
- v0y is the initial vertical velocity (which can be found using the initial velocity and the launch angle, but you haven't provided the angle)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

However, since the angle of projection is not given, we cannot calculate the initial vertical velocity. Please provide the launch angle of the projectile.