Exercise session #02 - 10/03/2021 - 3/10/2021 A tennis player launches the tennis ball vertically downward with an initial speed v₁ = -√98 and an initial height y₂ = 1.5m. The ball hits the ground and bounces back with a perfectly elastic collision. Find the time it takes for the ball to return to the initial height yo. For 0≤t≤tb v(t)= v₁- gt y(t) = y.+vt - gt² 0 = y₁ + v₁ t₁ - £gt ² to = ± √2+24 Lval (√₁+ 248-4) v₂ = V₁ - gtb = V₁² √ 1+242 <0 = > W 1vbl. MOY مر to O t M V₂ = -|v₂| 1D V₂ <0 1650 ܝܘ 2
Exercise session #02 - 10/03/2021 - 3/10/2021 A tennis player launches the tennis ball vertically downward with an initial speed v₁ = -√98 and an initial height y₂ = 1.5m. The ball hits the ground and bounces back with a perfectly elastic collision. Find the time it takes for the ball to return to the initial height yo. For 0≤t≤tb v(t)= v₁- gt y(t) = y.+vt - gt² 0 = y₁ + v₁ t₁ - £gt ² to = ± √2+24 Lval (√₁+ 248-4) v₂ = V₁ - gtb = V₁² √ 1+242 <0 = > W 1vbl. MOY مر to O t M V₂ = -|v₂| 1D V₂ <0 1650 ܝܘ 2
Related questions
Question
I don’t understand the last line of vb how they did the factoring method of v0 I don’t get what happened inside the radical
![Exercise session #02 - 10/03/2021-3/10/2021
A tennis player launches the tennis
ball vertically downward with an
initial speed v₁ = -√98 and an
initial height y₂ = 1.5m. The ball.
hits the ground and bounces back with
a perfectly elastic collision. Find the
time it takes for the ball to return
to the initial height yo.
For 0≤t≤tb
v(t) = V₁ - gt
y(t) = y + xt-gt²
0 = y₁ + v₁ t₁ - £gt₂²
to = ± √ 4 24+24
=
2,
V₁ = V₁- gtb = V₁² √ 1+ 2409
<O
Ivbl.
Vol
0
Vo
to tr
t
Yo
M f
1D
V₂ <0
450
V₂= |v₂|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F55cc6b9f-9ecf-4cea-b6f6-26c154ea6cfd%2Funz3dpe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise session #02 - 10/03/2021-3/10/2021
A tennis player launches the tennis
ball vertically downward with an
initial speed v₁ = -√98 and an
initial height y₂ = 1.5m. The ball.
hits the ground and bounces back with
a perfectly elastic collision. Find the
time it takes for the ball to return
to the initial height yo.
For 0≤t≤tb
v(t) = V₁ - gt
y(t) = y + xt-gt²
0 = y₁ + v₁ t₁ - £gt₂²
to = ± √ 4 24+24
=
2,
V₁ = V₁- gtb = V₁² √ 1+ 2409
<O
Ivbl.
Vol
0
Vo
to tr
t
Yo
M f
1D
V₂ <0
450
V₂= |v₂|
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 7 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)