Question

Math - Based - Physics .... QUESTIONS.
(I am working the parts 1-6 on my own, if you want, please come and view and correct if anything)


A soccer ball is kicked with an initial horizontal velocity of 11 m/s and an initial vertical velocity of 17 m/s.

Answer 1

AND?

Answer 2

I got disconnected I apologize.
I will post parts of the problem will do them here on my own and if I need help i will look for it.

Answer 3

http://i.imgur.com/I0prCLo.png

here is the IMAGE
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1) What is the initial speed of the ball?

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2) What is the initial angle θ of the ball with respect to the ground?

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3) What is the maximum height the ball goes above the ground?

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4) How far from where it was kicked will the ball land?

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5) What is the speed of the ball 2.5 seconds after it was kicked?

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6) How high above the ground is the ball 2.5 seconds after it is kicked?

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Answer 4

` PART 1: `

The initial speed of the ball (lets denote it with u) is:

\(\large u=\sqrt{11^2+17^2}=20.24\color{blue}{\left({\rm m/s}\right)}\)

Answer 5

` PART 2:`

\(\large \theta={\rm Tan}^{-1}\left({\rm 17/11}\right)\approx 57.09\)

Answer 6

Sorry i dont
know.

Answer 7

Now for part 3, we know that maximum height is reached when v=0.

Also, I am given:

a = -9.81m/s\(^2\)
u = 20.24m/s

So will I use? \(v^2=u^2+2ax\)

NOTE:
"u" denotes initial speed.
"x" denotes displacement/distance (in this case same thing)

\(v^2=u^2+2ax\)
\(0^2=17^2+2(-9.81)x\)
\(-17^2=2(-9.81)x\)
\(17^2=2(9.81)x\)
\(x=17^2/[2(9.81)]\approx 14.73\)

Answer 8

`PART 4`
How far from where it was kicked will the ball land?

`(Horizontal Displacement) = (Time) × (Horizontal Velocity)`

I know the horizontal velocity, but the time is unknown.
We can use the vertical components to find the time!!

Since the ball travels in a parabolic motion it is just twice the time
from v=17 (initial horiz velocity) to v=0 (the maximum height point, or the vertex)

so we went from 17 m/s to 0 m/s
how many second would that have took if every second we lose 9.81m/s?
17(m/s) / 9.81(m/s^2) ≈ 1.732 seconds

Now remember we go twice from v=17m/s to v=0m/s
So we really have 3.464 seconds of FULL motion time when ball is in air.

units work out too!

Now we can find the time.
`(Horizontal Displacement) = (Time) × (Horizontal Velocity)`
`(Horizontal Displacement) = (3.464) × (11)`
`(Horizontal Displacement) = 38.104`

Answer 9

I am having trouble with part 5.

Answer 10

A soccer ball is kicked with an initial horizontal velocity of 11 m/s and an initial vertical velocity of 17 m/s.

5) What is the speed of the ball 2.5 seconds after it was kicked?

Answer 11

v=u+at

u - initial speed
t - time
a - accel due to gravity = -9.81m/s^2

v=20.24+(-9.81)2.5=-4.285
speed is always positive, so 4.285

because we want to know what happens with speed at t=2.5

No, that doesn;t work for some reason.

Answer 12

11+(9.81)2.5=35.525
and that is incorrect either (I put it in my practice thingy and it denied)

Answer 13

oh, 20.24 not 11, I suppose.
initial velocity V< not speed of horiz.

Answer 14

20.24+(9.81)2.5=44.675

but that is too large. and I check it is not right