38.2 m/s 5.50 m 50.0° A golfer hits an approach shot with a velocity of 38.2 m/s at 50 degrees above the horizontal. The golf ball lands directly next to the hole, 5.50 m above the ball's original position, and stops immediately. 8. How long was the ball in the air?

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Chapter1: Units, Trigonometry. And Vectors
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Vo
38.2 m/s
5.50 m
50.0°
A golfer hits an approach shot with a velocity of 38.2 m/s at 50 degrees above the horizontal.
The golf ball lands directly next to the hole, 5.50 m above the ball's original position, and stops
immediately.
8. How long was the ball in the air?
Transcribed Image Text:Vo 38.2 m/s 5.50 m 50.0° A golfer hits an approach shot with a velocity of 38.2 m/s at 50 degrees above the horizontal. The golf ball lands directly next to the hole, 5.50 m above the ball's original position, and stops immediately. 8. How long was the ball in the air?
Expert Solution
Step 1

Given that,

The speed of a golfer, v = 38.2 m/s

It shot at an angle of 50 degrees above the horizontal.

The golf land directly next to the hole 5.5 m above the ball's original position. 

 

We need to find the time for the ball in the air. We can use the second equation of kinematics to find it.

h=voyt+12at2

Where

v0y is the vertical component of velocity

a = acceleration, a = -g (acceleration due to gravity)

 

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