A golf ball is struck with a five iron on level ground. It lands 85.0 m away 4.60 s later. What was the magnitude and direction of the initial velocity

It depends upon the initial "launch angle" of the golf ball, and the "launch velocity". The time spend coming down before it hit the ground was 2.3 s. The vertical velocity component was therefore Vyo = g*2.3s = 22.56 m/s. The horizontal velcoity component (which remains constant) is Vx = 85m/4.6s = 18.48 m/s. The "launch angle" is arctan 22.56/18.48 = 50.7 degrees

The initial velcoity is sqrt[(Vyo^2 + Vx^2]

To find the magnitude and direction of the initial velocity of the golf ball, we can use the equations of motion for projectile motion.

Let's break down the problem:

- The golf ball is struck on level ground, so the initial vertical velocity is 0 m/s. We only need to consider the horizontal component of the initial velocity.
- The golf ball lands 85.0 m away, which is the horizontal displacement.
- The time taken for the golf ball to land is 4.60 s.

Using the equation for horizontal displacement (assuming constant horizontal velocity):
displacement = velocity × time

We can rearrange this equation to solve for the initial horizontal velocity:
velocity = displacement / time

Plugging in the values:
velocity = 85.0 m / 4.60 s = 18.48 m/s

So, the magnitude of the initial velocity is 18.48 m/s.

Since there is no vertical component to the initial velocity, the direction of the initial velocity is purely horizontal. Therefore, we can say that the direction is straight ahead or parallel to the level ground.