A ball is thrown up in the air with a velocity of 35m/s. (Assume the ball starts with an initial height of 2m

Question

A ball is thrown up in the air with a velocity of 35m/s. (Assume the ball starts with an initial height of 2m)
At what time is it at an altitude of 1.0 meters above the ground? (Use the quadratic formula)

Answer 1

s = s(zero) + v(zero)t + (1/2)at^2
1 = 2 + 35t - 4.9t^2
4.9t^2 - 35t - 1 = 0
t = (-B +/- √[B62 - 4AC])/2A
t = (35 +/- √[35^2 - 4(4.9)(-1)]/9.8
t = 35 +/- 35.2789)/9.8
Lets take the positive answer.
t = 7.1713s

Answer 2

7.168 secs