A projectile is fired and lands t = 14 s later on the side of a hill, at a location 1400 m away horizontally and 420 m above its starting point. Find the projectile's initial horizontal velocity vx0 and initial vertical velocity vy0. Assume g = 10 m/s2.

68.6 m/s

To find the projectile's initial horizontal velocity (vx0) and initial vertical velocity (vy0), we can use the kinematic equations of motion.

First, let's find the time it takes for the projectile to reach the vertical distance (420 m) above its starting point. We can use the equation:

h = vy0*t - (1/2) * g * t^2

Where h is the vertical distance, vy0 is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.

Plugging in the given values:
420 = vy0*14 - (1/2) * 10 * (14)^2

Simplifying the equation, we get:
420 = 14vy0 - 980
14vy0 = 1400
vy0 = 100 m/s

Now, we can find the initial horizontal velocity (vx0). We know that the horizontal distance covered is 1400 m and the time taken is 14 s. We can use the formula:

d = vx0 * t

Where d is the horizontal distance and vx0 is the initial horizontal velocity.

Plugging in the given values:
1400 = vx0 * 14

Solving for vx0, we get:
vx0 = 100 m/s

So, the projectile's initial horizontal velocity (vx0) is 100 m/s and the initial vertical velocity (vy0) is 100 m/s.