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Vertical Motion Problems. MA.912.A.7.8 Use quadratic equations to solve real-world problems. . Vertical Motion Formula d=rt – 5t 2. 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.
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Vertical Motion Problems MA.912.A.7.8 Use quadratic equations to solve real-world problems.
Vertical Motion Formulad=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.
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.)
d=rt – 5t2 Maximum Height Object distance Ground
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
Graph Clink on link for graphing calculator. http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
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?
A football is kicked into the air with an initial upward velocity of 25 m/sec. • d. When will the ball hit the ground?
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.
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?
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?
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?
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.
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.
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.
3. • c. At what time does the ball reach its highest point? How high is the ball above the gym floor? Time when it goes In the basket. Time Thrown The ball reaches its highest point halfway between the time it is thrown and the time it reaches the basket.
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.