# A golfer is teeing off on a 170.0 m long par 3 hole. The ball leaves with a velocity of 40.0 m/s at 50.0 o to the horizontal. Assuming that she hits the ball on a direct path to the hole, how far from the hole will the ball land (no bounces or rolls)? [9.38 m]

1. Projectile motion is influenced only by gravity.
2. Range of projectile is given as [V^2 sin2(theta)]/g.
3. Using this equation, range is 160.62 m.
4. This is 9.38 m (= 170 - 160.62) short of the hole. This is an example of projectile motion. A projectile can be thought of as an object which is thrown with a certain velocity and whose motion is then only governed by the force of gravity. The projectile can be thrown vertically or at an angle like the golf ball or a cricket ball.

The horizontal distance traveled by a projectile is termed as the Range and is given by the following equation:

Range = `(V^2 sin 2theta )/g`

where V is the initial velocity with which the projectile is thrown, `theta` is the angle with the horizontal made by the projectile and "g" is the acceleration due to the gravity.

Here, the initial velocity is 40 m/s, angle made with the horizontal is 50 degrees, and using a value of 9.81 m/s^2 for acceleration due to gravity on the surface of Earth, we can calculate the range of the golf ball as:

Range = [40^2 sin 2(50)]/9.81 = 160.62 m

Thus, the golf ball (assuming no bounce or roll) will land 160.62 m from the point where it started from. Since the hole was 170 m away from the starting point, the ball land a distance of 170 - 160.62 = 9.38 m short of the hole.

Other parameters, such as the maximum height reached, time of flight, etc. can also be calculated by using the appropriate equations of projectile motion.

Hope this helps.