A ball is thrown directly downward with an
initial speed of 3.08 m/s from a height of
14.6 m.
After what interval does the ball strike
the ground? The acceleration of gravity is
9.8 m/s
2
To find the time it takes for the ball to strike the ground, we can use the equation of motion for vertical motion:
s = ut + (1/2)at^2
where:
s = displacement (in this case, 14.6 m)
u = initial velocity (3.08 m/s)
a = acceleration (gravity, -9.8 m/s^2, considering downward as negative)
t = time (unknown)
Rearranging the equation, we get:
t^2 - (2u/a)t - (2s/a) = 0
Substituting the values we have:
t^2 - (2 * 3.08 / -9.8)t - (2 * 14.6 / -9.8) = 0
Simplifying further:
t^2 + 0.628t + 2.98 = 0
Now, we can solve this quadratic equation to find t.
A quadratic equation can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation:
a = 1
b = 0.628
c = 2.98
Substituting these values into the formula, we get:
t = (-0.628 ± √(0.628^2 - 4 * 1 * 2.98)) / (2 * 1)
Calculating the expression under the square root:
√(0.628^2 - 4 * 1 * 2.98) = √(0.394384 - 11.92) = √(-11.525616) = undefined
Since the expression under the square root is negative, it means that there are no real solutions to this equation. This implies that the ball never strikes the ground.