a object is launched into the air vertically from ground level with initial velocity 70 m/s. What is the objec?

max. height is 249.7m.

Note that the equation of the height (in terms of m/s) is h(t) = -4.9t² + vot + h0, where t is time, vo is initial velocity and h0 is the starting position.  The problem gives us:

h(t) = -4.9t² + 70t

Then, by differentiation:

h'(t) = -9.8t + 70

Remember: h'(t) = v(t).

When the object is at max. height, h'(t) = 0.

0 = -9.8t + 70
70 = 9.8t
t = 70/9.8
t ? 7.14

Finally, substitute that value for h(t) to determine the answer!

h(7.14) = -4.9(7.14)² + 70(7.14)
h(7.14) ? 250 meters

Hence, the maximum height is approximately 250 meters high.

***Maximum Velocity***

Note that there is no max. velocity for the equation since h'(t) is linear.

I hope this helps!

height is given by    h =ut --1/2gt^2
  max ht = u^2/2g
        = 70^2/'2 x9.8
 ===  250 m