Question

please part B

4. The height and speed of a projectile launched with a speed of vi at an angle w is given by: h(t) = vyt sing - gt? v(e) = //-zvgtsing + { g27% The projectile will strike the ground when h(t)=0 which gives the time to hit tat2(11/8) sino. Assume o = 30°, v = 40 g = 9.81 m/s?; (a) By plotting h(t) and (t) on the same graph for Ost < 12 (s), determine the times when (1) the height is no less than 15 m and speed is simultaneously not greater than 36 m/s and (ii) the height is no less than 15m (8) Using MATLAB relational and logical operators, write a program to find the same times as in 4(a), and compare both the answers estimated from plots, and the program. (81 (6) When a constant voltage was applied to a certain motor initially at rest, its rotational speed (t) versus time was measured, as following table: Time(s) 1 2 3 4 5 6 7 8 11 Speed (rpm) 1210 1860 2280 2550 2710 2875 2882 2910 2930 Determine whether the following function can describe the data. If so, find the values of the constants y, and a (1) - v. (-e") [Data plot and the fitted plot must be shown) 191

Answer 1

• The graph is plotted using excel
• The given equation traces the curve of v(t) vs t, however a correction of 10% should be added for accuracy at t > 5

The given equation is an exponential equation. To check if the relation between v(t) and t is exponential, graph between the given two variables need to be plot S. No. t(s) 1 1 v(t) (rpm) 1210 1860 2280 2 2 3 3 4 4 2550 5 5 2710 6 6 2875 7 7 2882 8 8 2910 9 11 2930 3500 2882 2910 2930 3000 2875 2500 2710 2550 2280 2000 1860 1500 v(t) (rpm) 1210 1000 500 0 0 4 6 00 10 12 N t(s) From the graph we can observe that that v(t) is increasing exponentially with increase in value of t. Hence to represents the given variables mathematically, exponential equation should be used.

The given equation Vz Voc1-ext) com be verified by substituting the values of graph in the equation → by fubstituting vt) 1210 & tal we get, 1210 = Vo (1-0%) - (0) -> by substituting v (t) 2 1860 & tz 2 we get, 18602 Vo(1-6-28) --Lii) on dividing in by ci) we get, 1:5371z 1-e24 l-ea 1.5571: (+€*)(1-04) (1-e-a) 1:53712 itea so, put a in &z 0-6215 ci we get , Voz 2614.1705

so, le 0.0215 and Vo =261401705 the equation form will be -0.6215xt V(L) z 2014.1705 (1-é 0.625 8+) put t235 we get V (t) z 2209.04 which is very close to vt)2 2280 so, the given equation follows the graph of (t) v/s t Note: - as the value of time increases the error in value of ve) also increase so a correction of +200 for tzs