Question

Colby is standing on a bridge and using a catapult to throw rocks of the bridge into a lake. Suppose the function rr models the relationship between the height above the lake (in feet) of any rock thrown by the catapult in terms of the number of seconds elapsed since the rock was launched.

a. At 5:08pm Colby launches rock A into the lake. It takes 15 seconds to reach the water. Represent the height of rock A above the lake (in feet) 3 seconds after Colby launched rock A into the lake. 415 Preview b. Represent the height of rock A above the lake (in feet) t seconds after Colby launched rock A into the lake. 15-t)a Preview c. Suppose Colby launches another rock, rock B, with the catapult at 5:10pm. Complete the table below. # seconds elapsed since Colby launched Rock A 120 121 134 # seconds elapsed since Colby launched Rock B Preview Preview Preview d. Represent the number of seconds elapsed since Colby launched rock B given that t represents the number of seconds elapsed since Colby launched rock A. Preview syntax error e. Represent the height of rock B above the lake (in feet) 125 seconds after Colby threw rock A. 10a Preview f Represent the height of rock B above the lake (in feet) 12 seconds after Colby launched rock B. 1/5 Preview

Answer 1

Equations for projective motion x = vt cos theta y = vt sin theta - 1/2 gt^2 } s^rightarrow = -gj^Now T = 2v sin theta/g (a) Lapse starts at: 5: 08 pm time to reach river(T) = 15 sec Now v sin theta = 15g/2 = 15 times 9.8/2 = 73.5 m/s Now, height attain by stone- y_1 = (v sin theta)t - gt^2/2 t = 3 sec = 73.5 times 3 - 9.8/2 times 9 = 176.4 m = 578.74 ft (b) y_2 = 1(v sin theta)t - g/2 t^2)m = (73.5 t - 9.8/2 times t^2) times 3.28 ft y_2 = (241.08)t - 16.072 t^2 ft \r\n (e) second through would start at 5: 10 pm so, 125 second means rightdoublearrow 5 sec because v sin theta = 73.5 m/s y_3 - (v sin theta)t - 1/2 g t^2 = ((73.5) times 5 - 1/2 times 9.8 times 25) times 3.28 ft = 803.6 ft (f) t - 12 second y_4 = (N sin theta times 12 - 1/2 times 9.8 times 144) times 3.28 = 578.59 ft