Answer:
Option 2nd is correct
2 seconds does it take to hit the ground
Step-by-step explanation:
The equation of projection is given by:
\(s(t) = -at^2+v_ot+h\)
where,
\(a \approx -16 m/s^2\)
\(v_0\) is the initial velocity
and h is the initial height.
As per the statement:
A ball is dropped from a treetop 64 feet above the ground.
Initial height: s(0) = 64 feet
Since, initial velocity \(v_0 = 0\)
then equation becomes:
\(s(t) =-16t^2+0t+64\)
To find the how long does it take to hit the ground.
Substitute s(t) = 0 then ;
\(0 =-16t^2+0t+64\)
⇒\(0 = -16t^2+64\)
⇒\(16t^2= 64\)
Divide both sides by 16 we have;
\(t^2 = 4\)
Simplify:
\(t =\pm \sqrt{4} =\pm 2\)
Since, t is the time cannot be in negative
⇒t = 2 seconds
Therefore, 2 seconds does it take to hit the ground.
