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

Ground

distance

Object

Maximum Height

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

a. Write the related equation.

b. Calculate the height after 2 sec & 3 sec

€

d = 25t −5t 2

€

d = 25(2) −5(2)2

= 50 − 20

= 30m

€

d = 25(3) −5(3)2

= 75 − 45

= 30m

Graph

Clink on link for graphing calculator.http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

€

y = 25t −5t 2

1. 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?

€

20 = 25t −5t 2

5t 2 −25t +20 = 0

5(t 2 −5t + 4) = 0

5(t −1)(t − 4) = 0

€

t =1sec

or

t = 4 sec

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

d. When will the ball hit the ground?

€

0 = 25t −5t 2

5t 2 −25t = 0

5t(t −5) = 0

5t = 0 or t −5 = 0

€

t = 0 sec

or

t = 5 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.

a. Write the related equation.

€

d =15t −5t 2

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?

€

d =15(2) −5(2)2

= 30 − 20

= 10 m

€

d =15(4) −5(4)2

= 60 − 80

= −20 m

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?

€

0 =15t −5t 2

5t 2 −15t = 0

5t(t − 3) = 0

5t = 0 or t − 3 = 0

€

t = 0 sec

or

t = 3 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.

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

€

−50 =15t −5t 2

5t 2 −15t −50 = 0

5(t 2 − 3t −10) = 0

5(t +2)(t −5) = 0

€

t = −2 sec

or

t = 5 sec

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.

a. Write the related equation.

€

d = 6t −5t 2

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.

€

d = 6(0.3) −5(0.3)2

= 1.8 − 0.45

=1.35 m above basket

€

1.35 + 3

= 4.35 m

above 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.

€

0 = 6t −5t 2

5t 2 −6t = 0

t(5t −6) = 0

t = 0 or 5t −6 = 0

€

t = 0 sec

or

t =6

5sec =1.2 sec

3.

c. At what time does the ball reach its highest point? How high is the ball above the gym floor?

€

d = 6t −5t 2

€

t = 0 sec or t =1.2 secTime 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.

€

t = 0.6 sec

3.

c. At what time does the ball reach its highest point? How high is the ball above the gym floor?

€

d = 6t −5t 2

The ball reaches its highest point halfway between the time it is thrown and the time it reaches the basket.

€

t = 0.6 sec

€

d = 6(0.6) −5(0.6)2

d = 3.6 −1.8

d =1.8 m

€

The ball reaches

amaximum height

of 4.8 m.