(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A ball is thrown at 60 degrees above x-axis. It remains in the air for 2.9 seconds. What is the final displacement in the x (range) if the ball was thrown with the same initial velocity, but at an angle of 45 degrees above x-axis instead?

2. Relevant equations

y = (Voy * t) + (1/2 * ay * t^2)

Voy = Vo * sin(theta)

Vox = Vo * cos(theta)

x = 1/2(Vox + Vx)t

3. The attempt at a solution

So plugging into the first equation I get:

y = (Voy * t) + (1/2 * ay * t^2)

0 = (Voy * 2.9) + (1/2 * -9.8 * 2.9^2)

Voy = 14.21

Taking the result into the second equation I get the initial velocity:

Voy = Vo * sin(theta)

14.21 = Vo * sin(60)

Vo = 16.41

Am I on the right track so far? From here I would think I'd take it into the first equation again with the 45 degree angle so:

y = (Voy * t) + (1/2 * ay * t^2)

0 = (16.41sin(45)*t) + (1/2 * -9.8 * t^2)

0=11.60t-4.9t^2

t = 2.37

So I've got the time, and Vox from Vox = Vo * cos(theta) since Vo is the same for both throws.

For the equation

x = 1/2(Vox + Vx)t

(substitute Vox = Vo * cos(theta))

x = 1/2(16.41cos(45) + Vx)2.37

All I'm missing is Vx, I am lost where to go from here.

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# Homework Help: Projectile Motion - given angle, time; need displacement

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