Velocity The object will lose 98 ms of velocity every second because of gravity We can refer to them as gravity bites Position Use kinematics equation 2 to calculate the position of the object at any time ID: 783796
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Slide1
Vertical Launch
Physics
Slide2Let’s use kinematics to analyze an object that has been launched straight up moving 29.4 m/s
Slide3Velocity
The object will lose 9.8 m/s of velocity every second because of gravity.
(We can refer to them as “gravity bites.”)
Slide4Position
Use kinematics equation 2 to calculate the position of the object at any time.
∆ x = ½ at
2
+ v
0
t
Slide5v
2
=
x
2
=
v
1
=
x
1
=
v0 =x0 =
v4 =x4 =v5 =x5 =v6 =x6 =
v3 = x3 =
9.8 m/s39.2 m19.6 m/s24.5 m29.4 m/s0 m
-9.8 m/s39.2 m-19.6 m/s24.5 m-29.4 m/s0 m
0 m/s
44.1 m
29.4 – 9.8
19.6 – 9.8
9.8 – 9.8
0 – 9.8
-9.8 – 9.8
-19.6 – 9.8
∆ x = ½ at2 + v0t = ½ (-9.8)(1)2 + 29.4(1)
∆ x = ½ at2 + v0t = ½ (-9.8)(2)2 + 29.4(2)
∆ x = ½ at2 + v0t = ½ (-9.8)(3)2 + 29.4(3)
∆ x = ½ at2 + v0t = ½ (-9.8)(4)2 + 29.4(4)
∆ x = ½ at2 + v0t = ½ (-9.8)(5)2 + 29.4(5)
∆ x = ½ at
2
+ v
0
t
= ½ (-9.8)(6)
2
+ 29.4(6)
Slide6v
2
=
x
2
=
v
1
=
x
1
=
v0 =x0 =
v4 =x4 =v5 =x5 =v6 =x6 =
v3 = x3 =
9.8 m/s39.2 m19.6 m/s24.5 m29.4 m/s0 m
-9.8 m/s39.2 m-19.6 m/s24.5 m-29.4 m/s0 m
0 m/s
44.1 m
THINGS TO NOTICE:
Path is symmetric.
Time Up = Time Down
Launch speed = impact speed
The velocity at he highest point = 0 BUT the acceleration is still -9.8 m/s
2
(turning counts as acceleration).
The position is always positive BUT velocity up = + and velocity down = -
Slide7Symbolic Alphabet Soup
Solving for Height
TOP:
∆
x =
v
0
=
v =
a =
t =
Solving for Time
END:
∆x =v
0 =v =a =t =
0 mlaunch- launch- 9.8 m/s
2ttotalUse to find ttotal at the END
h
launch
0 m/s
- 9.8 m/s
2
t
up
= ½ of ttotalUse to find height at the TOP
Slide8NUMERIC EXAMPLE
A soccer ball is kicked straight up moving 15 m/s. How high will it go? How much air time will it have?
END:
∆
x =
v
0
=
v =
a =
t =
0m
15 m/s
- 15 m/s- 9.8 m/s2wantUse 1st Equation: v = at + v0
Solve for t:t = v – v0 a= -15 – 15 -9.8=
3.06 s
TOP:∆x =v0 =v =a =t =
want15 m/s0 m/s-9.8 m/s2½ (ttotal) =
½ (3.06) = 1.53 s
Use 3rd Equation:
∆x = ½ (v0
+v) t
Plug in and solve:
∆x = ½ (15 +0) (1.53)
=
11.5 m
Slide9PRACTICE
What launch speed is needed to keep a tennis ball in the air for 3.4 seconds if it is hit straight up?
Slide10PRACTICE
How long will an arrow stay in the air if it is fired straight up going 52 m/s?
Slide11PRACTICE
How high up will a jumper go if they leave the ground moving 6.4 m/s?
Slide12PRACTICE
Bob is trying to throw his backpack over a tall fence. If the fence is 7 meters tall, then how much time will the backpack be in the air?
Slide13PRACTICE
How much launch velocity is needed in order for gymnast to jump 1.2 meters while doing a back flip?
Slide14PRACTICE
A rocket is launched in the soccer field. After 8 seconds, the rocket crashes back into the grass. How high did the rocket go?