Answer:
1.63 seconds
Step-by-step explanation:
To calculate the ball's speed at different times and find the time it reaches its maximum height, we can use the equations of motion for uniformly accelerated motion. In this case, the ball is thrown straight up, and we can consider its acceleration due to gravity (g) to be approximately -9.81 m/s² (negative because it acts downward). The initial velocity (u) is +16.0 m/s (positive because it's directed upward).
After 2.00 seconds (t = 2.00 s):
We can use the following equation to find the velocity (v1) at this time:
v1 = u + at
v1 = 16.0 m/s + (-9.81 m/s²) * (2.00 s)
v1 = 16.0 m/s - 19.62 m/s
v1 = -3.62 m/s
So, the ball's speed after 2.00 seconds is approximately 3.62 m/s, and it's moving downward.
After 4.00 seconds (t = 4.00 s):
Using the same equation:
v2 = u + at
v2 = 16.0 m/s + (-9.81 m/s²) * (4.00 s)
v2 = 16.0 m/s - 39.24 m/s
v2 = -23.24 m/s
The ball's speed after 4.00 seconds is approximately 23.24 m/s, and it's moving downward.
After 5.00 seconds (t = 5.00 s):
Again, using the same equation:
v3 = u + at
v3 = 16.0 m/s + (-9.81 m/s²) * (5.00 s)
v3 = 16.0 m/s - 49.05 m/s
v3 = -33.05 m/s
The ball's speed after 5.00 seconds is approximately 33.05 m/s, and it's moving downward.
To find the time (tmax) when the ball reaches its maximum height, we know that at the maximum height, the final velocity (vf) is 0 m/s. We can use the following equation:
vf = u + atmax
Since vf is 0 m/s at the maximum height, we have:
0 = 16.0 m/s + (-9.81 m/s²) * tmax
Solving for tmax:
-16.0 m/s = -9.81 m/s² * tmax
tmax = (-16.0 m/s) / (-9.81 m/s²)
tmax ≈ 1.63 seconds
So, the ball reaches its maximum height after approximately 1.63 seconds.