This equation is derived from the following:
at maximum height the velocity is zero.
y(current height)= y0 (initial height)+v0t+.5at^2
take the derivative
32t-v0=0 so vo=32t
Plug this in for the v0 value
16t2 − (32t)t + h = 0
h=49ft
-16t^2+49=0
Solve for t:
16t^2=49
4t=7
t=7/4
Plug in t and h to solve for v0
16t2 − v0t + h = 0 where h=49 and t=7/4
16(7/4)^2-v0 (7/4) +49=0
:P
How fast would a ball have to be thrown upward to reach a maximum height of 49 ft? [Hint: Use the discriminant of the equation
16t2 − v0t + h = 0.
2 answers
How fast would a ball have to be thrown upward to reach a maximum height of 49 ft? [Hint: Use the discriminant of the equation 16t2 − v0t + h = 0.]
2 answers