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**Vertical Motion Problems**

MA.912.A.7.8 Use quadratic equations to solve real-world problems.

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**Vertical Motion Formula d=rt – 5t2**

The formula d=rt (Distance = rate X time) works when the rate is constant. When something is thrown upward into the air, the rate varies. The rate gets slower and slower as the object goes up, then becomes negative as it comes back down again.

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d=rt – 5t2 t is the number of seconds since the object was thrown upward. d is its distance in meters above where it was thrown. r is the initial upward velocity in meters per second. (The rate when the object was first thrown.)

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d=rt – 5t2 Maximum Height Object distance Ground

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**A football is kicked into the air with an initial upward velocity of 25 m/sec.**

Write the related equation. Calculate the height after 2 sec & 3 sec Ask students what they think may be happening between 2 sec and 3 sec that would explain why both times yield the same height (distance above the ground)?

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**Graph Clink on link for graphing calculator.**

Analyze the graph of the equation. Ask students to estimate the maximum height that the ball will reach and how many seconds would have elapsed when the ball reaches the max height. Clink on link for graphing calculator.

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**A football is kicked into the air with an initial upward velocity of 25 m/sec.**

c. When will it be 20 meters above the ground?

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**d. When will the ball hit the ground?**

A football is kicked into the air with an initial upward velocity of 25 m/sec. d. When will the ball hit the ground? Explain to students that the height at ground level is 0 meters. Also explain that although the height is 0 meters at 0 seconds and at 5 seconds, the only correct answer is 5 seconds because at 0 seconds, the ball has not yet left the ground.

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**2. Suppose that you throw a rock into the air from the top of a cliff**

2. Suppose that you throw a rock into the air from the top of a cliff. The initial upward velocity is 15 m/sec. Write the related equation. Ask students what they think may be happening between 2 sec and 3 sec that would explain why both times yield the same height (distance above the ground)?

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**2. Suppose that you throw a rock into the air from the top of a cliff**

2. Suppose that you throw a rock into the air from the top of a cliff. The initial upward velocity is 15 m/sec. b. How high will the rock be above the cliff after 2 sec? Where will it be after 4 sec? Have students think about the height of the rock at 4 seconds, ask what they think the negative might represent as it relates to this problem. You may want to go back one slide and have them look at the diagram in the previous slide.

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**2. Suppose that you throw a rock into the air from the top of a cliff**

2. Suppose that you throw a rock into the air from the top of a cliff. The initial upward velocity is 15 m/sec. c. When will it again be at the same level you threw it?

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**d. When will it hit the water, 50 meters below where you threw it?**

2. Suppose that you throw a rock into the air from the top of a cliff. The initial upward velocity is 15 m/sec. d. When will it hit the water, 50 meters below where you threw it? Analyze with students what the -2 seconds would represent and why it does not answer the question being asked.

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**3. A basketball player shoots a long shot**

3. A basketball player shoots a long shot. The ball has an initial upward velocity of 6 m/sec. When it is released, the ball is at the same level as the basket which is 3 meters above the gym floor. Write the related equation.

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**3. A basketball player shoots a long shot**

3. A basketball player shoots a long shot. The ball has an initial upward velocity of 6 m/sec. When it is released, the ball is at the same level as the basket which is 3 meters above the gym floor. b. After 0.3 seconds, how high is the ball above the basket? How high above the gym floor.

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**c. Assuming that the aim is good, when will the ball go in the basket.**

3. A basketball player shoots a long shot. The ball has an initial upward velocity of 6 m/sec. When it is released, the ball is at the same level as the basket which is 3 meters above the gym floor. c. Assuming that the aim is good, when will the ball go in the basket. Ask students what they think may be happening between 2 sec and 3 sec that would explain why both times yield the same height (distance above the ground)?

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3. c. At what time does the ball reach its highest point? How high is the ball above the gym floor? Time Thrown Time when it goes In the basket. The ball reaches its highest point halfway between the time it is thrown and the time it reaches the basket.

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3. c. At what time does the ball reach its highest point? How high is the ball above the gym floor? The ball reaches its highest point halfway between the time it is thrown and the time it reaches the basket.

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Chapter 2, Part 3 September 7, 2005. A ball is thrown downward (not dropped) from the top of a tower. After being released, its downward acceleration.

Chapter 2, Part 3 September 7, 2005. A ball is thrown downward (not dropped) from the top of a tower. After being released, its downward acceleration.

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