This is a question off of quest homework for UT physics I Initially (at time t =

Question

This is a question off of quest homework for UT physics I

Initially (at time t = 0) a particle is moving vertically at4.8 m/s and and horizontally at 0 m/s. The particleaccelerates horizontally at 1.3 m/s 2 . Theacceleration of gravity is 9.8 m/s 2 . At what timewill the particle be traveling at 35? with respect to thehorizontal?

Answer in units of s.

So I set up the equation tan -1[(4.8t-4.9t2)/.65t2] = 35 from the kinematicequation distance=x0+v0t+.5at2

2nd step

(4.8t-4.92)/(.65t2) = tan 35

Eventually, I found the time to be .89633 and my answer was off. Ilooked at the graph and at t=.89633 it actually had a negativeslope and didn't represent 35 degrees with respect to thehorizontal. So I'm guessing you can't set up trig on a nonlinearparametric function.

Explanation / Answer

I set this up similar to how you did as well. tan 35o = vy/vx 0.7 = (v0y + gt)/(vox +at) 0.7 = (4.8m/s +-9.8m/s2 * t)/(1.3m/s2 *t) Then rearranging to solve to t: 0.7(1.3m/s2*t) = 0.91m/s2 * t = (4.8m/s+-9.8m/s2 * t) 10.71m/s2 t = 4.8m/s t = 0.4482s Hope that helps

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.