Ghassan+Albuhairan

A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. The acceleration of gravity is 9.81 m/s2. a) What was the pelican’s initial speed? b) If the pelican was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?


 * Given: Y-Yo= 5.4m g=-9.8 X-Xo= 8m Voy= 0m/s RECALL: Y-Yo=Voy*t+1/2*a*t^2 -Using the above equation and the given values we can solve for time. -With the time we are able to adapt the equation for the X component and solve for the initial speed of the X component. - And with that we can solve for the fishes initial speed. Part 2: Take the speed that was previously calculated, and replace the old Y-Yo with 2.7 RECALL: Y-Yo=Voy*t+1/2*a*t^2 1. Using the above equation you can calculate the new time. 2 with that we can find the new X-Xo using the initial speed. 3. The X-Xo that you get is the answer**




 * Week 7:**

__**Week7:**__

01 . An object is hung from a spring balance attached to the celing of an elevator cab. The balance reads 65 N, when the cab is standing still. What is the reading when the cab is moving upward, (a) with a constant speed of 8 m/s and

Given: a=0 m/s², T=65N. we use this equation to solve it: F=T= ma + mg

(b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s²

Given: a = -2.4 (cuz it's deceleration) T = mg + m(-24) T= 49N. we use the same equation to find it: T = mg + ma