Question

A golf ball is launched at an angle of 45°. The ball has an initial vertical velocity of 12 m/s and an initial horizontal velocity of 12 m/s. Forget air resistance. The ball reaches its maximum height 1.2 seconds after it is launched and lands at the same height that is launched from. The total horizontal distance reached by the golf ball is about

a) 14.4 meters
b) 17.3 meters
c) 24.0 meters
d) 28.8 meters

Answer 1

The golf ball will land approximately 28.8 meters away from its launch point when projected at a 45° angle with an initial horizontal velocity of 12 m/s, considering a total time of 2.4 seconds in the air.

Step-by-step explanation:

The problem provided describes a classic case of projectile motion in physics without air resistance. To find the total horizontal distance (range) the golf ball traveled, we rely on the information given about its initial horizontal velocity and the time it took to reach its maximum height. Since the angle of launch is 45°, and the ball spends 1.2 seconds to reach its peak, it will take another 1.2 seconds to come down to its original level, making the total time in the air 2.4 seconds. To find the range, we multiply the total time by the initial horizontal velocity:

Range = Horizontal velocity × Total time

Range = 12 m/s × 2.4 s = 28.8 meters

Hence, the ball will land approximately 28.8 meters away from its launch point, making the correct answer (d).