Answer:
1.78 m/s
Explanation:
Distance to putt = 8 m
Distance the ball stops to the putt = 1.5 m
therefore distance traveled by the ball = 8 - 1.5 = 6.5 m
The ball stops at this point 1.5 m from the putt, therefore its final velocity at this point = 0 m/s
the ball was struck with an initial velocity of 1.6 m/s
Using the equation
[tex]v^2[/tex] = [tex]u^2[/tex] + 2as
where v is the final velocity of the ball = 0 m/s
u is the initial speed of the ball = 1.6 m/s
a is the acceleration of the grass
s is the distance the ball travels = 6.5 m
substituting values, we have
[tex]0^2[/tex] = [tex]1.6^2[/tex] + 2(a x 6.5)
0 = 2.56 + 13a
13a = -2.56
a = -2.56/13 = -0.197 m/s^2
If this acceleration of the grass is assumed to be constant, then to the initial speed needed to make the putt will be calculated from
[tex]v^2[/tex] = [tex]u^2[/tex] + 2as
where
v is the final speed at the putt = 0 m/s
u is the initial speed with which the ball is struck = ?
a is the acceleration of the grass = -0.197 m/s^2
s is the distance to the putt = 8 m
substituting values, we have
[tex]0^2[/tex] = [tex]u^2[/tex] + 2(-0.197 x 8)
0 = [tex]u^2[/tex] - 3.152
[tex]u^2[/tex] = 3.152
u = [tex]\sqrt{3.152}[/tex] = 1.78 m/s
