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A ball is thrown straight up at 16.0 m/s. Ignore the effects of air resistance. What is the ball's speed v; after 2.00 s? Vj = m/s What is the ball's speed v2 after 4.00 s? U2 = m/s What is the ball's speed uz after 5.00 s? U3 = m/s At what time tmax does the ball reach its maximum height?

User EBarr
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1 Answer

3 votes

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.

User Nicholas Green
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