April 8th, 2018, 11:32 AM  #1 
Newbie Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0  Help Me Kin Assignment
I need help with question 3...I have no idea how to solve it..

April 8th, 2018, 12:26 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575  projectile motion in 2 dimensions ... (a) $\Delta y = v_0 \sin{\theta_0} \cdot t  \dfrac{1}{2}gt^2$ $\Delta y = 35 \, m$, $v_0 = 120 \, m/s$, $\theta_0 = 30^\circ$ solve the resulting quadratic for $t$ (b) using $t$ found in part (a) ... $\Delta x = v_0 \cos{\theta_0} \cdot t$ (c) $v_{fx} = v_0 \cos{\theta_0}$ ... velocity in the $x$ direction remains constant using $t$ found in part (a) ... $v_{fy} = v_0 \sin{\theta_0}  gt$ $\theta_f = \arctan\left(\dfrac{v_{fy}}{v_{fx}}\right)$ 
April 8th, 2018, 12:49 PM  #3 
Newbie Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0 
3a) I got two answers as I used the quadratic formula, time cannot be negative so the answer I got is t=0.29s b) I got 5.36m so I rounded off to 5.4m c)1.429 degrees = 1.43 degrees 
April 8th, 2018, 01:03 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 
seems your calculator is in radian mode ... change to degrees

April 8th, 2018, 01:28 PM  #5 
Newbie Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0 
OOpsss...thank u 3a)12.29s=12.3s b)1278.25m=1278.3m c)0.58 degrees = 0.6 degrees Last edited by mayak201; April 8th, 2018 at 01:30 PM. 
April 8th, 2018, 01:33 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 
$0 = 35 + 60t  4.9t^2$ $a=4.9$, $b=60$, $c = 35$ $\dfrac{b  \sqrt{b^24ac}}{2a} \approx 12.8 \,sec$ (b) should be $\approx 1330 \, m$ (c) should be $\approx 32^\circ$ 

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