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 = V₁² + 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) = V₁ + at v² = 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 V f = 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 = V₁² + 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) = V₁ + at v² = 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 V f = 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

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?

for question 4 use the image labeled 4 to solve the problem. For question 5 use the image labeled 5 to solve.

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 = V₁² + 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)
= V₁ + at
v² = 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
V f = 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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