A ball is thrown upward with an initial speed of
24.5 m/s. When is it 19.6 m high? (two answers)
The formula is
s = sO + vOt + at2/2
Since its initial speed is 24.5 m/s upward, sO = 24.5
Since we assume it starts on the ground, sO = 0
Since we want to know when its distance is 19.6 m,
s = 19.6
Since we know that the acceleration is due to
gravity, and that the ball is thrown up on Earth,
(and not on the moon or Mars), then a = g = -9.8 m/s2.
So a = -9.8.
Plug these into
s = sO + vOt + at2/2
19.6 = 0 + 24.5t + (-9.8)t2/2
19.6 = 24.5t - 4.9t2
4.9t2 - 24.5t + 19.6 = 0
Clear of decimals by multiplying through by 10,
that is, by moving the decimals one place to the
right in all the terms.
49t2 - 245t + 196 = 0
Divide all terms by 49
t2 - 5t + 4 = 0
Factor the trinomial on the left sides:
(t - 4)(t - 5) = 0
Setting the first factor = 0
t - 4 = 0
t = 4 seconds
Setting the second factor = 0
t - 5 = 0
t = 5 seconds
So it reaches the height of 19.6 m after 4 seconds
(on its way up) and again after 5 seconds (on its
way down).
Edwin
