Projectile Motion golf ball

Sesner09

Hello, im having trouble with the question below..

1. A golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. A) How much time does the ball spend in the air? b) what is the longest "hole in one" the golfer can make if the ball doesn't roll?

2. right triangle, T=(Vo-V)/g, for b i know the eqn is x= Vx*t

3. i made a right triangle with the right angle on the bottom right. i made the hypotenuse 30.3 m/s. I believe the equation to find time is T=(Vo-V)/g. How would i find the time?

thank you

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rock.freak667

Homework Helper
use

$$y=y_0 +v_y t -\frac{1}{2}gt^2$$

at what angle must the golfer hit the ball to get the maximum range?

Sesner09

it doesn't give an angle, would y(the height) be 0 since it is not changing elevation?

rock.freak667

Homework Helper
it doesn't give an angle, would y(the height) be 0 since it is not changing elevation?
yes it would so your equation is y=vyt-(1/2)gt2 and you want to find when y=0

also it said it hit it to the maximum range, so there is only one angle the max range can be achieved, do you know what this angle is?

Sesner09

the only angle it gives is 90°, so 90?, i got 90 from the triangle that i drew..
the problem that i keep running into is that i get 0=30.3 m/s(t) - (1/2)(-10 m/s^2)(t^2)..... and that gives me t=0

Last edited:

rock.freak667

Homework Helper
the only angle it gives is 90°, so 90?, i got 90 from the triangle that i drew..
the problem that i keep running into is that i get 0=30.3 m/s(t) - (1/2)(-10 m/s^2)(t^2)..... and that gives me t=0
if the range is given by

$$R=\frac{v^2 sin 2\theta}{g}$$

for what angle is R maximum for?

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