A baseball is seen to pass upward by a window 25 m above the street with a verti

Question

A baseball is seen to pass upward by a window 25 m above the street with a vertical speed of 9 m/s. The ball was thrown from the street. What was its initial speed? m/s What altitude does it reach? m How long after it was thrown did it pass the window? s After how many more seconds does it reach the street again? s A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.6 s later. If the speed of sound is 340 m/s, how high is the cliff? m A stone is dropped from the roof of a high building. A second stone is dropped 2.50 s later. How far apart are the stones when the second one has reached a speed of 13.0 m/s? m

Explanation / Answer

9. a)using,v^2-u^2=2gs

=> 9^2-u^2=-2*9.8*25

=> u^2 = 81+490

u = 23.9 m/s

b)v at max height =0

max height=u^2/2g

= 571/(2*9.8)

= 29.13 m

c)time take to reac window

= v-u/(-g)

= (9-23.9)/(-9.8)

= 1.52 s using v=u+at a=-g in upward motion

d) time take to reach max height from window= 0-9/-9.8 = 0.918 s

time taken to reach ground from window= 0.918+0.918+1.52 =3.356

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10.

Acceleration due to gravity is equal to 9.81m/s.

Delta(x) = V0t + 0.5ata^2 Since we are starting from rest at the top of the cliff, V0 = 0

Delta(x) = 0.5(9.81)ta^2 ta represents the time it takes the rock to reach the water

Delta(x) = 4.905ta^2

Now we work with the time it takes once the rock hits the water.

V = Delta(x)/tb tb is how long it takes the sound to travel up the cliff

Delta(x)= 340/tb

Delta(x) = 0.5Vtb we combine formulas for uniform acceleration.

340/tb = 0.5(340)tb

tb = 1.41 s                    Now we subtract the time it took for the sound of the rock to travel back up from

4.6 - 1.41 = 3.19 s        the total time

Delta(x) = 4.905(3.19)^2 And plug that in for ta

Delta(x) = 49.91 m

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11.

T1 = time it takes the second stone to attain a velocity of 13 m/s

g = acceleration of gravity = 9.81 m/s^2

S1 = distance traveled by the second stone

T2 = the time of travel of the first stone = T1 + 2.5

S2 = distance traveled by the first stone

T1 = V/g = 13/9.81 = 1.325 s

S1 =(1/2)g(T1)^2 = (1/2)*9.81*(1.325)^2 = 8.61 m

T2 = T1 + 2.5 = 1.325 + 2.5 = 3.825 s

S2 = (1/2)g (T2)^2 = (1/2)*9.81*3.825^2 = 71.76 m

S2 - S1 = 71.76 - 8.61 = 63.15 m

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