A ball is thrown at the edge of a cliff 90.0m high at 42m/s. Calculate A.) the time required to hit the ground below. B.) the velocity. C.) the velocity when the ball is 60.0m above the ground
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Expert's answer
2010-10-10T17:57:44-0400
Let's denote the horizontal projection as vx and vertical projection as vy. The equations of the motion would be: X: vx = v0x;Y: vy = v0y - gt; h = h0 + v0yt - gt2 / 2. A. For our case: X: vx = v0x= 42; Y: vy = v0y - gt = 0 - 10t = -10t; 0 = 90 + 0 - 10 t2 /2 = 90 - 5t2. 90 = 5t2; t= √ 18 = 3 √ 2 sec = 4.24 sec; B. Let's find Y-projection instantly before hitting the ground: vy = -10t = - 10 * 3 √ 2 = -30 √ 2 m/sec; Using Pythagoras' theorem, V = √ vy2 + vx2 = √ 422 + 302*2 = 59.7 m/sec; C. Analogically to the previous: h = h0 + v0yt - gt2 / 2.60 = 90 - 5t2; 5t2 = 30; t = √ 6; vy = -10t = - 10 √ 6 ; V = √ 422 + 600 = 48.62 m/sec; Answers: A. t(h=0) = 4.24 sec. B. V(h=0) = 59.7 m/sec. C. V(h=60) = 48.62 m/sec.
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