The planet Zorg has an acceleration of gravity that is 10.0 times that of Earth,

Question

The planet Zorg has an acceleration of gravity that is 10.0 times that of Earth, and a negligible 5) a. b. 5) atmosphere, so that there is no air friction. A Royal Zorg Trooper spots an octomorph in a water tree. He points his zipbow at the monster and shoots a quantum arrow at him with an initial speed of 105. m/s. However, the instant the zipbow is fired, the octomorph drops straight out of the tree, into the water below a) The octomorph is initially h-120. m above the Royal Trooper's zipbow, and d 110, m away from him horizontally. When the warrior shoots his zipbow, the quantum arrow follows the parabolic path shown in Fig.-5b. At what height, ', should the Royal Trooper aim his zipbow so that the arrow will hit the octomorph before the octomorph falls into the water as it's falling from the tree? b) At what height, H, above the Royal Trooper will the octomorph be struck with the arrow? h' IH Figure 5- (a) The octomorph is in the tree h 120. m above the Royal Trooper's zipbow, and d 110. In away from him horizontally. (b) The Royal Trooper aims his zipbow at a height, h', and fires his zipbow as the octomorph drops out of the tree. The arrow hits the octomorph at a height, H, above the Royal Trooper. Page 5 of 6

Explanation / Answer

5) a = 10 x 9.81 = 98.1 m/s^2

in horizontal, t = d / v0 cos(theta)

t = 110/ (105 cos(theta))

cos(theta) t = 1.0476 ..... (i)

time taken by to drop,

( h - H) = 98.1 t^2 / 2

120 - H =49 t^2

in vertical,

H = (105 sin(theta)) t - 98.1 t^2 /2

H = 105 sin(theta) t - 120 + H

105 sin(theta) t = 120 ... (ii)

(ii)/(i) => tan(theta) = 120 / 110

theta = 47.5 deg

tan(theta) = h' / d

h' = 110 tan47.5 = 120 m . .......Ans

(b) cos(theta) t = 1.0476

t = 1.551 sec

120 - H = 49.05 x 1.551^2

H = 2.10 m

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