Projectiles.

Forums:
Email this Topic • Print this Page

Dear Forum Experts....Pls help me with the answer.
A ball is thrown with a velocity 25 m/s at an angle"x". A gust of wind blew horizontally making an acceleration on the ball at 5 m/s^-2. If the ball is to return to the throwers hands what would be the angle "x" be?
Thanks.

Topic Stats
Top Replies
Link to this Topic

Type: Question • Score: 0 • Views: 969 • Replies: 20

No top replies

View best answer, chosen by senakaS

@senakaS,

They put me on the night shift and I'm only allowed to answer yes or no, true or false, and multiple choice questions.

1 Reply

@Ponderer,

Yeah, but nobody's watching at this time of the morning.

0 Replies

Hi guys.....Hope you are not making a mock out of it! pls help.

1 Reply

@senakaS,

They did say I could ask questions. Was there any spin on the ball? At what point ( either time or distance ) did the gust of wind occur?

0 Replies

No spin, and time/distance is not given. The five answers are;
(1) tan^-1 0.5, (2) tan^-1 0.25, (3) tan^-1 2, (4) tan^-1 1 (5) tan^-1 0.2

2 Replies

@senakaS,

Let's see if I've got this right. Someone throws a ball at a given angle and velocity. The mass and circumference ( and thus surface area and drag) are not specified. Both factors would have an effect on the arc of travel. At some point
( also not given) a gust of wind blows the ball back to the thrower's hand.
You might try the Jet Propulsion Laboratory.

0 Replies

Angle is what needed in the question? The initial velocity and the acceleration on the ball given. Drag is not considered. Only the acceleration by the wind would make the ball return to the thrower.

0 Replies

@senakaS,

Assuming x is the angle to the horizontal,
the vertical component of velocity is 25 sinx. The time of flight would be given by v=u-gt where v=0
whence t=g/(25 sinx) crucial step

horizontally use equation s=ut-0.5 at^2 a=0.5 and s=0
so 0=25 cosx. g/25 sinx -0.5 . o.5 (g/25 sinx)^2
so g/tanx=0.25 (g/25sinx)^2

I'll let you find x ! Very Happy

0 Replies

@senakaS,

EDIT
To sum up.
Resolve vertically to get t, using v=0
NBI think I should have used 2t above being the time to max height and back.
Resolve horizontally using 2t, and using s=0

1 Reply

@fresco,

Thank you fresco......Excellent.

0 Replies

@senakaS,

Quote:

I'm only allowed to answer yes or no

I'm only allowed to answer them my Alzie's lets me

Fres thanks for Aka'S sake, and I wonder how long that took ya

1 Reply

@dalehileman,

It took me 10 mins plus 5 for correction.
I did not finally solve for x and would probably have tried a tan (x/2) substitution if the last equation got tricky

1 Reply

@fresco,

Quote:

10 mins plus 5

Still that constitutes a considerable effort, for which you're to be applauded

Quote:

I did not finally solve for x

That's probly not reqd anyhow

1 Reply

@dalehileman,

I agree tan(x/2) is not needed .
In fact my result (prompted by you) was tan x=g/a
But this suggests a much simpler solution... that of a triangle of perpendicular vectors.... since force =mass x accn... with g as the vertical and a the horizontal.

1 Reply

Great follow up.....thanks guys.

0 Replies

@fresco,

Quote:

.... simpler solution... .... triangle of perp... vectors.. f =mass x accn... g .... vertical and a ...horiz....

Wow but I'm impressed by you guys.

No really, I'm perfectly serious. I look back at my own ed. with the utmost scorn...

0 Replies

s=ut+1/2at^2

Vertically,
0 = 25sinθt + 0.5(-10)t^2
0 = 5t (5sinθ - t)
t = 5sinθ

Horizontally,
0 = 25cosθt + 0.5(-5)t^2
0 = 5t (5cosθ -0.5t)
t = 10cosθ

Combining,
5sinθ = 10cosθ
sinθ/cosθ = 10/5
tanθ = 2

θ = arctan 2

1 Reply

@ascribbler,

Vertically t=(2v sin x)/g
Horizontally 0=v cos x. t - 0.5 a t^2
Dividing by t
v cos x= a (v sin x)/g
So tan x = g/a = 20 ish not 2

Note that the velocity v is irrelevant. All that matters is the balance of vertical and horizontal forces which depend on accn.

0 Replies

Mmm.

The hypotenuse (the resultant vector) is 25 at angle θ

The adjacent side ( the horizontal x axis vector) is 5 , the ball travels at 5 along the x axis and returns to the origin due to horizontal wind 5

The opposite side ( the vertical axis vector) is Square root (25^2 - 5^2) Pythagoras

Tangent = Opposite / Adjacent = 4.89898

Tan^-1 (4.89897) = θ = 78.463 degrees

1 Reply

Forums
» Projectiles.

Read Question
Reply to All

Projectiles.

Quick Links

My Account

able2know