Q: How high does the stick go and What is the speed of the stick when it hits the ground if A stick held at a height of 2.50 meters is thrown into the air at 2.00 meters per second?

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We have no idea how big the rock is, and no way to figure it out. But we can calculate that it reaches 11.48 meters above the ground before it starts falling.

They should reach the ground together, since their initial vertical speed is the same, namely zero.

The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2

At the time the ball is thrown, which is "time 0" the downward speed is 40 m/s.Each second, the downward speed will increase by 9.8 m/s.1. Work out the speed at the end of the first second. This will be 49.8 m/s.2. Then work out how many meters it would have gone in the first second.3. Now work out the ball's height. This is the height at "time 1".4. Draw the ball at time 0 and time 1 on a sheet of paper to help you think.Now, repeat steps 1-4 until the ball's height is close to 0 or goes past 0. Your current "time X" will tell you how many seconds went by for it to get that far.

The ball was thrown horizontally at 10 meters per sec, and the thrower's arm was 78.4 meters above the base of the cliff.

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It depends on the height of the building and also on the direction the object is thrown in (up, down etc.).

We have no idea how big the rock is, and no way to figure it out. But we can calculate that it reaches 11.48 meters above the ground before it starts falling.

They should reach the ground together, since their initial vertical speed is the same, namely zero.

The height of the ball when it rebounds is 5 meters, as given in the question.

The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2

An object thrown vertically up wards from the ground returned back to the ground in 6s after it was thown up if it reached a height of 12m calculate?

At the time the ball is thrown, which is "time 0" the downward speed is 40 m/s.Each second, the downward speed will increase by 9.8 m/s.1. Work out the speed at the end of the first second. This will be 49.8 m/s.2. Then work out how many meters it would have gone in the first second.3. Now work out the ball's height. This is the height at "time 1".4. Draw the ball at time 0 and time 1 on a sheet of paper to help you think.Now, repeat steps 1-4 until the ball's height is close to 0 or goes past 0. Your current "time X" will tell you how many seconds went by for it to get that far.

The answer depends on whether the ball is thrown vertically upwards or downwards. That critical piece of information is not provided!

Using the horizontal distance and horizontal velocity of the ball, you can calculate the time it took for the ball to reach the ground. Then, you can use the time and the vertical acceleration due to gravity to find the height of the cliff using the equation: height = (1/2) * gravity * time^2. In this case, the height of the cliff would be approximately 11.3 meters.

The ball was thrown horizontally at 10 meters per sec, and the thrower's arm was 78.4 meters above the base of the cliff.

At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2

The maximum height of the ball above the ground can be calculated using the vertical component of the initial velocity. Assuming no air resistance, the formula to determine maximum height is h = (v^2 sin^2(theta)) / (2g), where v is the initial velocity (16 m/s), theta is the angle (40 degrees), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, you can find that the maximum height of the ball is approximately 14.1 meters.