Answer :
Answer:
It takes the rock t = 3.98 seconds to reach 3 meters from ground level
Explanation:
We need to set the kinematic equation for the vertical position of this motion in one direction with the zero (reference level) at ground level. So, the initial position of the rock is at positive 25 meters, and we need to find the time it takes to reach positive 3 meters.
Recall as well that the only acting acceleration is that due to gravity (g = 9.8 m/s^2) opposed to the object's initial velocity (Vi = 14 m/s). Then the position equation becomes:
[tex]y_f=y_i+v_i\,t-\frac{1}{2} \,9.8 \,t^2\\3 = 25 + 14\,t-4.9\,t^2\\4.9\,t^2-14\,t-22=0[/tex]
and we need to solve for "t" using the quadratic formula:
[tex]t=\frac{14+/-\sqrt{(-14)^2-4\,(4.9)(-22)} }{9.8} \\t=3.984 \,\,\,\, or\,\,\,\, t=-1.127[/tex]
Since we are looking for a positive value for time, starting when the rock is launched, we select the positive answer, which should come in seconds:
t = 3.98 seconds
(rounded to two decimals as requested)
Answer:
It takes the rock t = 3.98 seconds to reach 3 meters from ground level
Explanation: