Question

Submission 3 (0.66/1 points) Friday, July 24, 2020 07:15 PM +08 Suppose a small cannonball weighing 98 N is shot vertically upward, with an initial velocity Vo = 94 m/s. The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by d s/dt = -9 (equation (12) of Section 1.3). Since ds/dt = v(t) the last differential equation is the same as dv/dt = -g, where we take g = 9.8 m/s2. If air resistance is incorporated into the model, it stands to reason that the maximum height attained by the cannonball must be less than if air resistance is ignored. (a) Assume air resistance is proportional to instantaneous velocity. If the positive direction is upward, a model for the state of the cannonball is given by mdv = -mg - kv, where m is dt the mass of the cannonball and k> 0 is a constant of proportionality. Suppose k = 0.0025 and find the velocity v(t) of the cannonball at time t. v(t) = 39294e -0.00025t -39200 m/s (b) Use the result obtained in part (a) to determine the height s(t) of the cannonball measured from ground level. s(t) = 157176000(1 – e-0.00025t) – 392001 m Find the maximum height attained by the cannonball. (Round your answer to two decimal places.) 450.08 X m

Answer 1

-0'0002st S(t) = 1571760001-e 100025+) -39200 t metre Height the carmonball Let masumum is H -0.00025X9.58035799 H= 157176000/l-e - 39 doo 9.50035792 metre -0.00239500Gcdi H= 157176000 (1-e - 375550.03 metre H=157176000(0.00239222359)-375550.03 H = 450.097 metre H = 450.10 metre rounded to two decimal places B612

solution -0.000 art Vt) = 39294 e – 3.12oo mis Initial velocity cannon ball velocity of the Vo = 94 m/s After reaching marimum height velocity of the cannon batt cannon ball to time taken by the reach maximum height -0'000 art 39294e 39200 0.000ast 39294 3 9200 0 = e 0.00025t ln 39294 39 doo t = 0.0023950 89 48 0'000 25 t = 9.580357 9 2 sec 8612