Respuesta :
The distance d between the golfer and point b where the ball first lands is 4933.17 ft
Initial velocity = 250.0ft/s
Launch angle ([tex]\theta[/tex] )= 25°
slope angle([tex]\phi[/tex]) = 5°
Let d is the distance on the slope after which projectile lands
∴ horizontal distance covered by projectile
x = dcos[tex]\theta[/tex]
Vertical distance covered by projectile
y = dsin[tex]\theta[/tex]
now Considering horizontal motion
Initial horizontal velocity [tex]v_{x} = v cos25[/tex]
then position after time t
d cos5 = 250cos25×t
d = 227.44t ---------1.
when we consider vertical motion then
Initial vertical velocity [tex]v_{y} = v sin25[/tex]
[tex]v_{y} = v sin\theta t-\frac{1}{2}gt^{2}[/tex]
-d sin5 = 250 sin25.t - gt²/2
-d = 1212.25t - 56.221 t²
1212.25t - 56.221 t² + d = 0 ----2
put value of d in equation 2
1212.25t - 56.221t² + 227.44t =0
t(1212.25+227.44 - 56.221 t) =0
t(1234.69 - 56.221 t) = 0
t = 21.96 ,t≠0
substitute the value of t in equation 1
d = 227.44t
d = 227.44×21.96
d = 4933.17 ft
The distance d between the golfer and point b where the ball first lands is 4933.17 ft
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