Answer:
The value is [tex]v = 3.75 \ m/s [/tex]
Explanation:
From the question we are told that
The height of the ball above the ground is [tex]h = 1 \ m[/tex]
The angle is [tex]\theta = 37 ^o[/tex]
The distance of Alice from Bob is d = 50 m
The initial speed of ball is u = 20 m/s
Generally the height of the ball is mathematically represented as
[tex]h = -u sin(\theta)t + \frac{1}{2} (g) t^2[/tex]
=> [tex]1 = -20 sin(37)t + \frac{1}{2} * 9.8 t^2[/tex]
=> [tex]1 = -12t + 4.9 t^2[/tex]
=> [tex] 4.9 t^2 - 12t -1 = 0[/tex]
using quadratic formula we have that
[tex]t = 2.536 \ s[/tex]
The other value of time obtained is negative so we would not consider it as time can not be negative
Generally the distance the ball traveled in the horizontal direction is
[tex]s = ucos(\theta ) * t[/tex]
=> [tex]s = 20 * cos (37 )* 2.536 [/tex]
=> [tex]s = 40.5 \ m [/tex]
Generally the distance that Alice will travel before she get to the ball is
[tex]L = d - s[/tex]
=> [tex]L = 50 - 40.5 [/tex]
=> [tex]L = 9.5 \ m [/tex]
Generally the minimum speed of Alice is mathematically represented as
[tex]v = \frac{ 9.5}{ 2.536}[/tex]
[tex]v = 3.75 \ m/s [/tex]
