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Safari File Edit View History Bookmarks window Help webassign.net PRACTICE IT Use the worked example above to help you solve this problem. A ball is thrown from the top of a building with an initial velocity of 22.0 m/s straight upward, at an initial height of 51.1 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure. (a) Determine the time needed for the ball to reach its maximum height. (b) Determine the maximum height. (e) Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. Time Velocity m/s (d) Determine the time needed for the ball to reach the ground (e) Determine the velocity and position of the ball at 5.08 s. m/s Position EXERCISE HINTS: GETTING STARTED I TMSTUCK! A projectile is launched straight up at 51 m/s from a height of 83 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below. (a) Find the maximum height of the projectile above the point of firing. (b) Find the time it takes to hit the ground at the base of the cliff. (c) Find its velocity at impact. m/s Need Help? Read Your work in question(s) 2, 3 will also be submited or saved

Explanation / Answer

a] At the highest point, v = 0

u = 22 m/s

use v = u - gt

=> 0 = 22 - 9.8t

=> t = 2.245 s

b] in this time, the height covered is:

h = - [0 - 222]/2(9.8) = 24.694 m

this is the maximum height from the roof. Maximum height relative to the ground will be:

H = 24.694 + 51.1 = 75.794 m.

c] Time taken to reach the same initial height will be:

T = 2t = 4.49 s.

and using conservation of energy, the velocity at the initial height while the ball is coming down will be:

v = 22 m/s [downwards].

d]

v = u - gt

=> v = 22 - 9.8(5.08) = - 27.784 m/s

therefore the ball is travelling downwards.

and the position will be:

S = 51.1 + ut - (1/2)at2

=> S = 51.1 + (22)(5.08) - 4.9(5.08)2 = 36.408 m

this is the position above the ground.

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