 Physics Help Forum Projectile Motion Problem for Beginner Physics Student
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 New Users New to PHF? Post up here and introduce yourself! Mar 31st 2017, 07:05 AM #1 Junior Member   Join Date: Mar 2017 Location: NYC Posts: 4 Projectile Motion Problem for Beginner Physics Student I have a problem I can't solve. A stone is thrown from a building 45.0m high at an angle of 30 degree below the horizontal. It strikes the ground 57.0m away. A. What is the time of flight? B. What is the initial speed? C. What is the speed and angle of the velocity vector with respect to the horizon at impact? The answers are A. 1.57s B. 41.8m/s C. 51.3 and -45.0 degrees. I CANNOT figure out the time of flight as 1.51s. I've tried all the formulas I know. PLEASE HELP! This is driving me crazy! Last edited by nackel; Mar 31st 2017 at 09:53 AM.   Mar 31st 2017, 08:43 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 Start by setting up equations for x(t) and y(t). First for the horizontal position x(t): $\displaystyle x(t) = x_0 + v_{0x}t$ The initial horizontal velocity is equal to the stone's initial velocity times cos(30). Let's let T = the time the stone hits the ground - this yields: $\displaystyle x(T) = 57 = V_0 \cos(30) T$ (equation 1) Then for the vertical displacement: $\displaystyle y(t) = y_0 + v_{0y}t + \frac 1 2 a t ^2$ and at t=T: $\displaystyle y(T) = 0 = 45 - V_0 \sin(30)T -\frac 1 2 g T^2$ (equation 2) You now have two equations in two unknowns of V_0 and T. Solve for T. topsquark likes this.   Mar 31st 2017, 09:24 AM #3 Junior Member   Join Date: Mar 2017 Location: NYC Posts: 4 Forgive me, but I'm still confused. How can I solve for T if I don't know V_o? How can I solve for T if I don't know V_o?   Mar 31st 2017, 09:32 AM #4 Junior Member   Join Date: Mar 2017 Location: NYC Posts: 4 I used y=-1/2gt(squared) since Vyo=0 and got 3.02s. I just can't figure out what to do to get 1.57s. Last edited by nackel; Mar 31st 2017 at 09:53 AM.   Mar 31st 2017, 11:21 AM #5 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 Show me your work. If you sub V_0 from the first equation into the second and solve for T it should come out OK. Also, Vyo does NOT equal 0 - it equals -Vo sin(theta). Remember the projectile is thrown at a downward angle, so Vo does have an initial vertical component. topsquark likes this.   Mar 31st 2017, 02:47 PM #6 Junior Member   Join Date: Mar 2017 Location: NYC Posts: 4 My work is invalid because I was using 0 for Vyo. I see how right you are about the Vyo existing as the stone is thrown at a 30 degree down angle. I hope you can put up with my inexperience, but I can't see how to substitute Vo from the 1st equation into the second. I don't have a Vox and I don't know how to find it with only theta and delta x. This is why the question is driving me crazy. I feel like I don't have enough information to solve the problem, but since I'm new to this, I don't know what to do.   Mar 31st 2017, 06:20 PM #7 Senior Member   Join Date: Aug 2008 Posts: 113 $\Delta y = v_0 \sin{\theta} \cdot t - \dfrac{1}{2}gt^2$ Using $g = 9.8 \, m/s^2$ ... $-45=v_0 \sin(-30) \cdot t - 4.9 t^2$ $0=45 + v_0 \cdot \left(-\dfrac{1}{2}\right) \cdot t - 4.9 t^2$ $0=90-v_0 \cdot t - 9.8t^2$ $\Delta x = v_0\cos{\theta} \cdot t$ $57=v_0 \cos(-30) \cdot t$ $57=v_0 \cdot \dfrac{\sqrt{3}}{2} \cdot t \implies v_0 \cdot t = \dfrac{114}{\sqrt{3}}$ substitute $\dfrac{114}{\sqrt{3}}$ for $v_0 \cdot t$ in the $\Delta y$ equation and solve for $t$. Last edited by skeeter; Apr 1st 2017 at 09:42 AM.  Tags beginner, motion, physics, problem, projectile, student Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Jaded New Users 2 Mar 4th 2017 09:29 AM niniking Kinematics and Dynamics 3 Feb 19th 2011 05:11 PM pirateicechick Kinematics and Dynamics 3 Oct 7th 2010 10:09 PM Mattpd Kinematics and Dynamics 0 Sep 18th 2010 10:44 PM Rehaan_genius Kinematics and Dynamics 8 Aug 16th 2008 10:31 PM