VISUAL PHYSICS ONLINE PROBLEM P0113A

Transcription

1 VISUAL PHYSICS ONLINE PROBLEM P0113A Two balls are launched from the top of a cliff. Ball A has an initial velocit of 8.00 m.s at an angle of 30.0 o w.r.t. the horizontal and ball B has an initial velocit of 10.0 m.s at 60.0 o to the horizontal. The height of the cliff above sea level is 4.00 m. Ignoring air resistance, calculate: A. The initial components of the velocit of the two balls. B. The time taken for both ball to reach their maimum heights. C. The positions of the two balls (horizontal and vertical components w.r.t Origin) when the are at their maimum heights above sea level. What are the maimum heights of the balls above sea level? D. The speed of the balls at their maimum height. E. The relative position of ball B w.r.t. ball A when ball A is at its maimum height. F. The relative velocit of ball B when ball A is at its maimum height. G. The velocities of the balls as the enter the sea water. H. The flight times for the balls to enter the water.

2 I. The horizontal distance from the base of the cliff to where the balls enter the sea water. J. Sketch the following graphs (scaled aes) for both balls: (s / s), (s / t), (s/t), (v / t), (v / t). To help ou gain a better understanding of solving numerical phsics problems, ou should work through the Matlab code to see how to approach solving such problem. Even if ou don t know about Matlab, ou will be able to figure out how I solved the problem. Take note of the letters used to identif the phsical quantities.

3 % sp_projectiles.m clear all close all clc % INPUT ====================================== g = 9.81; u = [8 10]; A = [30 60]; h = 4; a = 0; a = - g; % Initial velocites u = u.* cosd(a) u = u.* sind(a) % When both balls are their highest positions % times, velocities and displacements th = -u./ a sh = u.*th * a * th.^2 sh = u.* th % When ball A is at its highest position: time th(1) % Position of ball B sa = sh(1); sa = sh(1); sb = u(2) * th(1) sb = u(2) * th(1) + 0.5*a*tH(1)^2 sba = sb - sa sba = sb - sa sba = sqrt(sba^2 + sba^2) angleba = atan2d(sba,sba)

4 % When ball A is at its highest position: time th(1) % velocit of ball B va = u(1) va = 0 vb = u(2) vb = u(2) + a * th(1) vba = vb - va vba = vb - va vba = sqrt(vba^2 + vba^2) anglevba = atan2d(vba,vba)

5 % As the balls enter the water s = -h vaw = u(1) vaw = -sqrt(u(1)^2+2*a*s) vaw = sqrt(vaw^2+vaw^2) angleaw = atan2d(vaw,vaw) vbw = u(2) vbw = -sqrt(u(2)^2+2*a*s) vbw = sqrt(vbw^2+vbw^2) anglebw = atan2d(vbw,vbw) taw = (vaw - u(1))/a tbw = (vbw - u(2))/a saw = u(1) * taw sbw = u(2) * tbw

6 Solution

7 A Initial velocities u ucos u usin Ball A u A 6.93 m.s u 4.00 m.s A Ball B u B 5.00 m.s u 8.66 m.s B B At the maimum height, the vertical velocit is zero v 0. Time to reach ma height v u a t v 0 a g t u a Ball A t 0.41 s Ball B t 0.88 s C At the maimum height for both balls: Horizontal displacement components w.r.t the Origin calculated from s u t Vertical displacement components w.r.t. to the Origin are calculated from s u t a t 1 2 2

8 Ball A s 2.83 m s 0.82 m Ball B s 4.41 m s 3.82 m The heights above sea level are Ball A h 4.82 m Ball B h 7.82 m D At the maimum heights, the vertical components of the velocit are zero. So, the speeds of the ball are equal to their initial horizontal velocities u A 6.93 m.s u 5.00 m.s B E The position of ball A when at maimum height is t 0.41 s s 2.83 m s 0.82 m A A The position of ball B at time t 0.41 s s u t s 2.04 m B s u t a t sb 2.72 m When ball A is at its highest position, the position of ball B w.r.t to ball A is Ball A s 2.83iˆ 0.82 ˆj Ball B s 2.04iˆ 2.72 ˆj A B

9 The relative position of ball B w.r.t. ball A is ˆ s s s i ˆj m BA B A s 0.79iˆ 1.90 ˆj m BA The displacement of ball B w.r.t ball A when ball A is at its highest position is s s s 2.06 m 2 2 BA BA BA BA s BA a tan 112 s BA o w.r.t. + X ais

10 F When ball A is at its height position t v v 0.41 s A 6.93 m.s A 0 m The velocit of ball B at time t 0.41 s v B u B 5.00 m.s vb ub a t vb 4.66 m.s Ball A v 6.93iˆ 0 ˆj A Ball B v 5.00iˆ 4.66 ˆj B The relative velocit of ball B w.r.t ball A is ˆ v v v i ˆj m BA B A v 1.93iˆ 4.66 ˆj m BA The velocit of ball B w.r.t ball A when ball A is at its highest position is v v v 5.04 m.s 2 2 BA BA BA BA v BA a tan 112 v BA o w.r.t. + X ais

11 G The velocities of the balls can as the enter the water can be found from the equations v u v u 2a s 2 2 Ball A v 6.93 m.s m.s 4.00 m.s 4 m u a u s v 9.72 m.s v m.s o Ball B v 5.00 m.s m.s 8.66 m.s 4 m u a u s v m.s v m.s o

12 H The time of flight can be calculated from v u a t t v u a Ball A Ball B t 1.40 s t 2.15 s I The range of the ball as the enter the water can be calculated from s Ball A Ball B u t s 9.69 m s m Note: in solving kinematics problems with constant equation ou never have to solve the quadratic equation s u t a t because ou can use a number of alternative equations v u a t 2 2 v u 2a s v u vavg s vavg t 2

13 J VISUAL PHYSICS ONLINE If ou have an feedback, comments, suggestions or corrections please Ian Cooper School of Phsics Universit of Sdne