A golf ball was hit and projected at an angle of 65° with the horizontal. If the initial velocity of the ball was 60m/s.Calculate the time the golf ball was in the air and the horizontal distance the ball travelled

Answer 1

The airborne time will be given by:

t = v1/g where t is the time taken, v1 is the initial velocity and g is the acceleration due to the gravity pulling the golf ball towards the center of the earth.

t=60/9.8
t=6.12244897959 seconds.

Multiplying that by 2 to have the total airborne time,

actual time = 12.2448979592 seconds.

Now, for the horizontal distance traveled,

Using simple height and distance, if the golf ball covered a distance of 60 meters in one second at an angle of 65° then the horizontal distance covered must be 25.3570957044 meters.

Multiplying that by the airborne time, we have,

25.3570957044 * 6.12244897959
= 155.247524721 meters.

So the golf ball was in the air for 6.1 seconds and the horizontal distance covered was 155.2 meters.