A ball is thrown directly downward, with an initial speed of 8.76 m/s, from a height of 30.8 m. After what interval does the ball strike the ground?
2006-10-01 15:57:41 · 3 answers · asked by warning9 2 in Science & Mathematics ➔ Physics
By the equations of motion:
s = ut + 1/2 at^2
where s = distance travelled
u = initial velocity
a = acceleration of particle
t = time taken for particle to travel
As the ball is being thrown directly downwards, gravity pulls it down, thus gravitational acceleration takes place in the movement of the ball. Assume in this case, that gravitational acceleration, g = 9.81 m/s^2.
30.8 = 8.76t + 1/2 (9.81t^2)
30.8 = 8.76t + 4.905t^2
4.905t^2 + 8.76t - 30.8 = 0
Solve this quadratic equation using the quadratic formula.
Then you get t = 3.53 (to 2 s.f.) and -7.11 (2 s.f)
But, since time, t, in this case, cannot be negative,
t = 3.53 (2 s.f.)
Therefore, the ball strikes the ground at 3.53 seconds, neglecting air resistence.
2006-10-01 16:17:52 · answer #1 · answered by chocho604 2 · 0⤊ 0⤋
as soon as it stops bouncing lol
2006-10-01 16:05:03 · answer #2 · answered by howcor 3 · 0⤊ 0⤋
30.8m Ha Ha
2006-10-01 16:05:05 · answer #3 · answered by fsjock 2 · 0⤊ 0⤋