A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground?
+14995 A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground?
\(mgh=\frac{1}{2}mv^2\)
\(\large h=\frac{v^2}{2g}\)
\(\Large h=\frac{\frac{12^2m^2}{s^2}}{2\times 9.81\frac{m}{s^2}}\)
\(\large h=7.339m\)
!
+9479 I think I can do this problem, but I'm not the best at these so it could be wrong.
Using 9.8 m/s2 as the approximation for gravity:
acceleration function: → h''(t) = -9.8
velocity function: → h'(t) = -9.8t + 12
position function: → h(t) = -4.9t2 + 12t
I didn't explain all the steps I did there because I don't know if my way of doing it is the best way; it's just the only way I know how to do it. So I don't want to confuse you any if my way is not the way you've been taught to do it.
We need to find the values of t where the velocity, h'(t) = 0. The velocity will be zero right at the max height.
0 = -9.8t + 12
-12/-9.8 = t
60/49 = t
So at t = 60/49, the velocity is 0. So let's find the position, h(t), when t = 60/49.
h(60/49) = -4.9(60/49)2 + 12(60/49)
h(60/49) = -4.9(3600/2401) + 720/49
h(60/49) = -360/49 + 720/49
h(60/49) = 360/49 ≈ 7.347 meters
+14995 A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground?
\(mgh=\frac{1}{2}mv^2\)
\(\large h=\frac{v^2}{2g}\)
\(\Large h=\frac{\frac{12^2m^2}{s^2}}{2\times 9.81\frac{m}{s^2}}\)
\(\large h=7.339m\)
!
