5 Throwing Upwards: Final Velocity Going down Givens: Man x= 1.82m or O t = 1.22 or 61 v₁ = 6 m/s (5,978) Vf= want a = -9.8m) 5² V top = 0-V ₂ O=V; Vi bits lets ignore the going up part, and turn it into a drop problem V₁ = V₂₁² + at V₁ = gt V₁ = (-9.8) (0.61) V₁ = -5.978 1/5 -6~/5 VF=-6m/s only half the story, use half the total time (vy-v₁) v = v₁ + at v = v² + 2ax 1 Ft + at² x = (" + "/^) ₁ 10 4 Throwing Upwards: Break down Vf. Time Going Up A tua = 1.225 Givens: x = 1.82m/ t = want up V₁ = 6 m/s O top a = -9.8 m/s² 2 Vf=V₁ +at 0= 6 + (²9.5) t 6 -5.8 = -9.8 t 1615 = tup Time Going Down Time total-Time up = time down 1.22 0.61 0.61 seconds to come down FOR ALL Projectiles that have a total vertical displacement of zero (they launch and land at same height) The total time in air is evenly split between the time up, and the time down. Max height is achieved at half of total time (vf - vi) t V = V₁ + at v² = v² + 2ax 1 x = v₁t + = at² 2 • (³₁ +01) ₁ t 2
5 Throwing Upwards: Final Velocity Going down Givens: Man x= 1.82m or O t = 1.22 or 61 v₁ = 6 m/s (5,978) Vf= want a = -9.8m) 5² V top = 0-V ₂ O=V; Vi bits lets ignore the going up part, and turn it into a drop problem V₁ = V₂₁² + at V₁ = gt V₁ = (-9.8) (0.61) V₁ = -5.978 1/5 -6~/5 VF=-6m/s only half the story, use half the total time (vy-v₁) v = v₁ + at v = v² + 2ax 1 Ft + at² x = (" + "/^) ₁ 10 4 Throwing Upwards: Break down Vf. Time Going Up A tua = 1.225 Givens: x = 1.82m/ t = want up V₁ = 6 m/s O top a = -9.8 m/s² 2 Vf=V₁ +at 0= 6 + (²9.5) t 6 -5.8 = -9.8 t 1615 = tup Time Going Down Time total-Time up = time down 1.22 0.61 0.61 seconds to come down FOR ALL Projectiles that have a total vertical displacement of zero (they launch and land at same height) The total time in air is evenly split between the time up, and the time down. Max height is achieved at half of total time (vf - vi) t V = V₁ + at v² = v² + 2ax 1 x = v₁t + = at² 2 • (³₁ +01) ₁ t 2
Related questions
Question
Use (g= -9.8 m/s/s)
What was the average total time of your tossed directly into the air object?
The average total time of my object tossed directly into the air object was 0.66 seconds.
Question 4:
How long does it take for the object to reach the maximum height? How long does it take to go from the maximum height to be caught again?
Question 5:
What is the final velocity of the object as it returns to the thrower's hand?
Use the image labeled 4 attached to solve exactly like the problem for question 4. Use the image labeled 5 to solve question 5, exactly like the problem.
Transcribed Image Text:5 Throwing Upwards: Final Velocity
Going down
Givens:
Man
x= 1.82m
or
O
t = 1.22 or 61
v₁ = 6 m/s (5,978)
Vf= want
a = -9.8m) 5²
V top = 0-V ₂
O=V;
Vi
bits
lets ignore the going up part, and turn
it into a drop problem
V₁ = V₂₁² + at
V₁ = gt
V₁ = (-9.8) (0.61)
V₁ = -5.978 1/5
-6~/5
VF=-6m/s
only half the
story, use half the
total time
(vy-v₁)
v = v₁ + at
v = v² + 2ax
1
Ft +
at²
x = (" + "/^) ₁
10
Transcribed Image Text:4
Throwing Upwards: Break down
Vf.
Time Going Up
A tua = 1.225
Givens:
x = 1.82m/
t = want
up
V₁ = 6 m/s
O
top
a = -9.8 m/s²
2
Vf=V₁ +at
0= 6 + (²9.5) t
6
-5.8
= -9.8 t
1615 = tup
Time Going Down
Time total-Time up = time down
1.22 0.61 0.61 seconds to come
down
FOR ALL Projectiles that have a
total vertical displacement of zero
(they launch and land at same
height)
The total time in air is evenly split
between the time up, and the time
down.
Max height is achieved at half of
total time
(vf - vi)
t
V = V₁ + at
v² = v² + 2ax
1
x = v₁t + = at²
2
• (³₁ +01) ₁
t
2
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