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the graph represents an object that is shot upward from a tower and then falls to the ground. The independent variable is time in seconds and the dependent variable is the object's height above the ground in meters. 1. How tall is the tower from which the object was shot?2. when did the object hit the ground?3. estimate the greatest height the object reached and then the time it took to reach that heightindicate this situation on the graph below by drawing on the graph given

the graph represents an object that is shot upward from a tower and then falls to-example-1
User Bddckr
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1 Answer

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14 votes

In the graph, the height, measured in meters, is on the y-axis and the time, measured in seconds, is on the x-axis. It shows the variation of the height of an object shot from a tower with respect to the time.

1. Since the shot was made from the top of the tower, the starting height of the object (time zero) is equal to the height of the tower. This is represented in the graph by the y-intercept of the function, that is, the value of y when x=0.

The y-intercept is at (0,10) which indicates that the height of the tower is 10m

2. The point where the object hit the ground, the height is zero meters, so in the graph, it is determined by the point where the parabola reaches the x-axis (y=0) known as the x-intercept.

The coordinates of the x-intercept are (6,0)

This indicates that the object hit the ground 6 seconds after being shot.

3. The variability of the object's height describes a parabola, where the vertex of the parabola indicates the highest point the object can reach.

You have to determine the coordinates of the vertex by looking at the graph. These are, approximately (2.8,93)

The y-coordinate of the vertex indicates the highest point of the object, which is 93meters.

The x-coordinate of the vertex indicates the time it took the object to reach said height, which is 2.8 seconds

User Stefan Vogt
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