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A golf ball is given an initial velocity of 38.2 m/s at an angle of 42.0o above the horizontal. How far away will the ball land on level ground?

Question

A golf ball is given an initial velocity of 38.2 m/s at an angle of 42.0o above the horizontal. How far away will the ball land on level ground?

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Step 1

The golf ball will be in a projectile motion. The maximum horizontal distance reached by the ball is the range of the ball. The equation for the range of the projectile motion is given by

sin 2e
R =
(1)
g
Here, R is the range, vo is the initial velocity, 0 is the initial angle of launch (angle with the horizontal)
and g is the acceleration due to gravity.
The value of g is 9.80 m/s2 .
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sin 2e R = (1) g Here, R is the range, vo is the initial velocity, 0 is the initial angle of launch (angle with the horizontal) and g is the acceleration due to gravity. The value of g is 9.80 m/s2 .

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Step 2

Substitute the numerical val...

(38.2 m/s) sin (2x 42°)
R =
9.80 m/s2
= 148.1 m
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(38.2 m/s) sin (2x 42°) R = 9.80 m/s2 = 148.1 m

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