Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
Answer:
2.11 seconds
Explanation:
We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):
![v_f=v_i-g*t\\-5.2 = 15.5 - 9.8\,t\\9.8 \,t= 15.5 + 5.2\\t = 20.7/9.8\\t = 2.11 \,\,sec](https://tex.z-dn.net/?f=v_f%3Dv_i-g%2At%5C%5C-5.2%20%3D%2015.5%20-%209.8%5C%2Ct%5C%5C9.8%20%5C%2Ct%3D%2015.5%20%2B%205.2%5C%5Ct%20%3D%2020.7%2F9.8%5C%5Ct%20%3D%202.11%20%5C%2C%5C%2Csec)
<span>The answer to this problem is magnesium. I hoped I helped someone with this</span>