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A glaucous-winged gull, ascending straight upward at $5.20 \mathrm{m} / \mathrm{s},$ drops a shell when it is 12.5 $\mathrm{m}$ above the ground. (a) What are the magnitude and direction of the shell's acceleration just after it is released? (b) Find the maximum height above the ground reached by the shell. (c) How much time does it take for the shell to reach the ground? (d) What is the speed of the shell just before it hits the ground?

(a) 9.8 $\mathrm{m} / \mathrm{s}^{2}$ downwards(b) 13.88 $\mathrm{m}$(c) 2.21 $\mathrm{s}$(d) 16.49 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 2

One-Dimensional Kinematics

Motion Along a Straight Line

Rutgers, The State University of New Jersey

University of Washington

Hope College

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Okay, So in this problem we have our lockers, Ringo extending a word with our work. Constant speed off. Let's see, crossed a speed of 5.2 5.2 meters per second. And we know that when they go reaches a maximum height off 12.5 meters, she drop is a shell. The problem once to know what was the magnitude and direction off the acceleration off the shelf? Well, we know they're physically when the shell detach off the girl she's too has the same velocity upwards off 5.2 meters per second. But no, she's under Ah, gravity deceleration Wolf G equals nine point 81 meters per second. Scratch and gravity's always points to the ground. So the answer to this problem, the first item is going to be a magnitude of deceleration 9.81 in the direction it's going to be down. Words. This is the answer for the first item on the second item the second I think we want to discover what was the maximum height off the shell we call X t Toto distance. She covers. Okay, well, we know that the shell is still moving upwards until she gets in a maximum height. And in this point she has a velocity zero in the initial point velocity or 5.2. So, after being dropped by the group, the wing of the Ingles, the show is too has velocity upwards still moving upwards until she reaches its maximum height. So since we already have velocities and we don't have time, we're just going to use the velocity as a function off space. Then we know the final velocity. Zero initial velocity. 5.2 square miners Because when gravity is in the opposite direction off their movement, this is gravity. We need to put a minor sign. So this is going to be miners two times nine point anyone. Delta X. So So we're going to have the difference. Inquisition it close acts, not acts. It goes to 27 points. Your foreign, divided by two times nine points, made you one, so this is going to be just one point 37 or that's round up to 38 tree eight meters. So since we started a movement from initial position roof sprinting here 12.5 meters, the total distance cover by the shell the maximal height is going to be 12.5 plus 1.38. So this is going to be 13 point eight eight meters. This is the answer from the second item. The third eye tonight and see, I don't see. We need to discover How much time does it take for the shell? Heats the ground. So how are we going to calculate this? We need to discover the time for this show reaches its maximum height. With this, we have our first time Then that time furnish show really from these maximum height unto the ground here. So this is the two movements with two different times that we need to calculate. So the first time is a movement with constant deceleration. And we just needs to use V equals zero minus g t miners. Because the gravity is opposite opposite to the movement. Okay, so the final velocity we know that he's your initial velocity. We know that it's 5.2 minus g d. Wait. We know the gravity. The gravity is just 9.81 nine point 81. I'm Steve. So the first time to this shell reaches his maximum height it's going to be That's pretty in here the first time. Do you want going to be 5.2? Divided by nine points. 81 This is equals two 0.53. A little boring 53 seconds And what is the second time? The second time is going to be accelerated movement, and we know that we also can use this same equation. But this time refer positive gravity acceleration because she's in the centre action of the movement. So let's see. Actually, we can use He's a creation because we don't know what was the final velocity off this shell when he reached the ground. So we need to use other equation. We're going to use the position as a function of time, the zero t plus G g to square. So we know here the initial velocity is here initial position and also be zero. So we have the total. The total distance we already calculated is 13.88 This is going to be 13.88 close nine point maybe one divided by two T square That's finished. This compilation in order page we have in Key Square equals two 13 point 88 times two Divided by nine point 81. This is two point 82 two point 82. So the square route off T it is just, let's see 1.60 eight, one point 68 seconds. So the total time of this movement on two she reaches the ground is going to be one point 68. Plus, that's a general 0.53 little 0.53. So this is something is equal to two point true point 21 seconds. This is a total time of the movement of the shell. Finally, the last item we need to discover what was the speeds off the shell just before hitting the ground. So now we can use this equation here because we have the time. And the only few minutes we just need to discover is the final velocity. So let's see, That's finding here the movement off the phone, we have a positive gravity is a celebration. We have initial velocity of zero, and the time of the fall is just 1.68. So this is going to be V equals nine points 80 watt times one point 68. So finally last answer. The last item of this problem is just 16 point 49 meters per second. This is the final answer to this problem. Thanks.

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