A ball is tossed vertically upward with an initial speed of 18.2 m/s. How long does it take before the ball is back on the ground? (Assume the ball is tossed upward from the ground.)
let g = 9.8 m/s^2
v = Vi - g t
0 at top
9.8 t = 18.2
t = 1.86 seconds upward
by symmetry, 1.86 seconds downward
t total = 3.72 s
Oh, you are doing calculus
a = -9.8
then
v = -9.8 t + constant
at t = 0, v = 18.2
so
v = 18.2 - 9.8 t
now height h
h = -(9.8/2)t^2 + 18.2 t + constant
at t = 0, h = 0
h = -4.9 t^2 + 18.2 t
when is h = 0?
t = 0 (of course)
and
t = 18.2/4.9 = 3.71 s close enough :)
To find the time it takes for the ball to be back on the ground, we need to consider the motion of the ball.
When the ball is tossed vertically upward, it will experience a constant acceleration due to gravity acting in the opposite direction to its motion. This acceleration is approximately 9.8 m/s², directed downward.
We can use the equation of motion:
`vf = vi + at`
where:
- vf: final velocity (which is 0 m/s when the ball reaches the top or lands on the ground),
- vi: initial velocity (18.2 m/s),
- a: acceleration (-9.8 m/s²),
- t: time.
When the ball reaches its maximum height, its final velocity will be zero. Therefore, we can use this equation to find the time taken to reach the maximum height:
`vf = vi + at`
`0 = 18.2 - 9.8t`
Solving this equation for t, we have:
`9.8t = 18.2`
`t = 18.2 / 9.8`
`t ≈ 1.86 s`
So, it takes approximately 1.86 seconds for the ball to reach its maximum height.
To find the total time it takes for the ball to be back on the ground, we need to consider that the time it takes to reach the maximum height is equal to the time it takes to fall back to the ground.
Thus, the total time it takes for the ball to be back on the ground is approximately 1.86 seconds.