Answer:
t = 3.25 seconds.
Explanation:
A person throws a baseball from a height of 7 feet with an initial vertical velocity of 50 feet per second is given by :
[tex]h=-16t^2+vt+h_0[/tex]
[tex]h=-16t^2+50t+7[/tex]
We need to find the amount of time the baseball is in the air before it hits the ground. ATQ,
h = 0
[tex]-16t^2+50t+7=0\\\\t=\dfrac{-50 \pm \sqrt{50^2-4\times (-16)(7)} }{2(-16)}\\\\t=\dfrac{-50 + \sqrt{50^2-4\times (-16)(7)} }{2(-16)}, \dfrac{-50- \sqrt{50^2-4\times (-16)(7)} }{2(-16)}\\\\t=-0.134\ s,3.25\ s[/tex]
Neglecting negative value, t = 3.25 s
So, the baseball will hit the ground at t = 3.25 seconds.
