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Question:

This is expanding on the angle of launch question. If a projectile is thrown 100m at an angle of 30 degrees, what is the v₀and y_max?

For v_0,I have gotten 100m = v²+ sin 2(30) / 9.8m/s
    then 100m X 9.8m/s = 980
    980m/s = (v²)(sin 60) = 1131.6 = v² or √1131.6 = 33.6 m/s

For y_max, I have tried to get t through v_o / g = 33.6 / 9.8m/s = 3.43 s,
    then plug into y = v₀+ v₀t + 1/2gt²,
    y = (33.6m/s) + (33.6m/s)(3.43s) - 1/2(9.8)(3.43s)²y = 91.2m height at peak

Does this look correct? Thanks for helping out!

Bonus Point Alert! Earn +4 additional karma points for helping this annual member.

Answers:

tejdhar

Apprentice

Karma Points: 104

Date Posted: 7/3/2008 11:02:46 AM Status: Live

Asker's Rating: None Provided Moderator's Rating: Somewhat Helpful

Response:

use the flma
R=V₀²Sin 2θ/g

limafoxtrot

Scholar

Karma Points: 200

Date Posted: 7/3/2008 11:05:54 AM Status: Live

Asker's Rating: N/A-Posted by Person Asking Question

Response:

What I do not understand is if the ball is launched at the same angle (30 degrees) but only travels 65m, the velocity decreases to 2.77m/s, which make sense, but I then get 140m for max height? Intuitively, I don't understand why the ball would achieve a greater height at the same launch angle and with less velocity? Anyone explain?

Kayen

Oracle

Karma Points: 21,761

(Ohio University Main Campus)

Date Posted: 7/3/2008 11:12:10 AM Status: Live

Asker's Rating: Lifesaver

Response:

Given R = x(max) =100 m; θ = 30^o; v_o = ?

(a) Youranswer is correct. But better follow this.

We know that R = (v_o²/g) sin 2θ

Therefore,

. v_o² = R.g/Sin 2θ = 100x9.8/0.866

= 1131.6

v_o = √1131.6 = 33.6 m/s

(b)You have found t and then y; It is ok But you used a wrong formula for y.

y = v₀+ v₀t + 1/2gt² is wrong; You can see it is dimensionally wrong also.

as lhs, y has dimension of length, whereas the first term on the right side, vo has dimension of L/T.

The correct formula is,

y = v₀t + 1/2gt²= 33.6x3.43 - 0.5x9.8x3.43x3.43

= 115.25 - 57.65 = 57.6 m

But it is easier and quiker to use, v² - u² = 2aS for motion in the y direction.

At the top, v = 0; v_o² = 2g.y_max

y_max = v_o² /2g = 33.6x33.6/2x9.8

= 57.7 m

limafoxtrot's Comment:

This was very helpful and prevented a major error on my part. Thank you so much!!

Kayen (Enjoys helping you uderstand..)

Please do not forget to Rate my Response.

zsm28

Oracle

Karma Points: 14,574

Date Posted: 7/3/2008 2:02:34 PM Status: Live

Asker's Rating: None Provided Moderator's Rating: Somewhat Helpful

Response:

initial velocity v = 33.6 m/s is right.
But the max height h = 57.6 m is wrong.
The formula h = vt - gt²/2 and t = v/g are both wrong.
It should be h = v_yt - gt²/2 and t = v_y/g, here v_y = vsin(30) = v/2 and t = v/(2g)
so h = (v/2)*v/(2g) - gv²/(8g²) = v²/(8g)
or directly use v_y² = 2gh, (v/2)² = 2gh
h = v²/(8g) = 33.6²/(8*9.8) = 14.4 m

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projectile motion question, need help.

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