Question

6. [10 points) For the following question, to model the free fall of a falling rock, assume the usual idealizing simplifications for solving "free fall" problems. Work with the approximate value g = 9.8 m/s2 for the Earth's gravitational acceleration. Consider the following experiment. The Leaning Tower of Pisa is known worldwide for its nearly four- degree lean. The height of the tower is 55.86 meters from the ground on the low side, and 56.67 meters on the high side. Let us assume that a person standing on the top of the tower on the low side throws a rock of mass m = 0.4 kg at time t = 0 seconds vertically upwards, releasing the rock from their hand at the exact height of 57 meters above ground with an initial velocity of 17 m/s. Because of the initial upwards throw, the rock first flies upwards towards its peak height before it starts falling downwards, falling past the person standing on the tower until it eventually hits the ground. Assume that the rock comes to an immediate rest" once it hits the ground. Let s(t) denote the function that describes the height of the rock above the ground during its vertical fall down to the ground, and let v(t) denote its velocity, and a(t) denote its acceleration at any given time to (a) Sketch a drawing of this experiment. (b) State the relevant facts from physics and calculus, and explain in a few English sentences how they lead to a second order differential equation for the function s(t). (c) State the differential equation that must be satisfied by the function s(t). Furthermore, state the initial conditions on 8(t) that describe the rock's initial height and its initial nonzero vertically upwards velocity. (d) Find the general solution (t) of the differential equation in c). (e) Find the solution (t) of the initial value problem in c). (f) Use your answer in e) to find the exact time when the rock hits the ground, and the velocity with which the rock hits the ground." (8) Use your answer in e) to find the exact time when the rock reaches its peak height, and the corresponding peak height of the rock during its free fall. 10 (h) Sketch the graph of the position function 8(t) over the time interval (0,6).

Answer 1

Page 1 57 56.67 55.86 s n (62 egh for the The acceleression of a body is determined by newton's law (F=ma)! We need paktion sth. Now acceleration a second order time derivative of position. Hence will have second order dilluential lah. we И ms (Alenterit tane) - mg Stro- 57 m initial anditints :- 17 mis vlt-0) = mg WH) att)= v= m w of do Johor g fat rits Vo = - gt 2 vlt = Vor gt

Page - ③ get SH). Integrate to वड म vo- gf sit Las ds - f (xo get let s its - So Vot - gt2 2 s26) Sot vot- 1 297² we hane, 17 mis mital value V = Vlt=o) = so = sltao) = 57m ga 9.8 slt) = 57 + 17t 9.8 2 Ist 57 17t- 496 (Position) Tvelocity also Tv (= 17- 9.8t Ruck Ints the szo ground se 62 57 + 17t- 4.9 +2=0 for t So he f= 4.962 17t-57 zo 17 ± √ (17)² + 4 (4-9 (67) 9.8 5.65 -2-15 tzo But 5.65 It

Page- 1 L relowly at t=5.65 is r(5-6)= 17. – 9.8)(5-6) 39 mis (-re sign implies downward velocity). Rock hits the ground after soos with a speed of 379 mls. 09 bat peak 179 8t t = TF 9.8 S 1-7 s hai Peak height is s(t = 1.7) s(t-1.7) = 57+ 17 (1.3) - 4.9 (1.772 71-7 m peak height of 71.7 m Rock reaches 1-7s after graph can be plotted with the eably help of si) 57 Chraph not to Scale. 1.75 5-65 n
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