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Question

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6
Standing on the top ledge of a 55 meter high building you throw a ball straight up with an initial speed of 31 m/s. How long, to the nearest second, does it take to hit the ground?
7
If the building in 6) was 50 meters high and you throw the ball at 30 m/s, how high to the nearest meter does it go?
8
A person jumps from a window 38 meters high and is caught in a firefighter's net which stretches 0.6 meter. To the nearest tenth of a m/s^2, what is the magnitude of the person's acceleration in the net?
9
How far to the nearest meter can a sprinter running at 9 m/s, travel in the time it takes a rock to fall 83 meters?
10
A helicopter travelling upward at 135 m/s drops a package from a height of 500 meters. To the nearest second how long does it take to hit the ground?

If you want to check your work, the answers are the folowing:
6) 8
7) 96
8) 620.7
9) 37
10) 31

Answer 1

6). Initial speed 31m/s.

When it reaches the ground its speed is down ward and hence it is (- 31) m/s,
(Since we take upward as positive in this case)

From v = u +at, t = v-u/ a.

Or from acceleration = change in speed / time, time = change in speed/ acceleration.

Therefore t = {31 - (-31)} / 10

..................t = 62/10

..................t= 6.2 OR 6 seconds.


7).From v^2 = u^2 + 2as,
Since v= 0, and a = -g, (Since we take upward as positive in this case)

s = - u^2 / (-2g) = u^2/ (2g)

s = 30 x30 / (2x 10) = 45m = 45m.


8).Assume that the net is near the ground.

Since he jumps down, a = g. And u=0. (Since we take downward as positive in this case)


Therefore, v^2 = 2gs

v^2 = 2x10x 38 = 760

v= 27.6m/s.

In the net his initial speed is 27.6 m/s, final speed is zero and distance traveled is 0.6m.

Acceleration is a.

From v^2 = u^2 + 2as,

..............0 = 27.6 + 2a x 0.6

...............a = -23 m/s^2. (Nearest tenth of a meter). Negative sign indicates that the acceleration is upward.


9) Initial speed of the rock =0,

Distance traveled = 83m

Acceleration = g.

s = (½) g t^2, since u= 0.

t^2 = 2x83 / 10 = 16.6.

t = 4.second.

Distance traveled by the sprinter during this time is 4. x 9 = 36 m


10) Initial speed = -135m/s. (since upward), (we take downward as positive in this case)
Acceleration = g = 10 (say)
Height= 500m.
s = ut + (1/2) a t^2
500 = - 135 t + 5 t^2.
100 = - 27t + t^2,
t^2 - 27t -100 = 0

b^2 - 4 a c =1129
Square root of 1129 = 33.6
The roots are (27 + 33.6) /2 and (27 -33.6) /2

30.3 second after start or

-3.3 second before start.

.

Answer 2

On 8 it cant be 620.7
your math could be right I didnt check.
1. How high is the net? 1 meter? 2? 10?
Anthing above 0 you would subtract that from the 38 meters of hight.
2. If the net is at the 38 meters below the window, then it is on the ground and can not strech .6 meters

Answer 3

i'll the first one...

no, the equation is in two bits:

(i.) when the ball is thrown vertically off the ledge, it reaches a zero velocity, and then returns to the original displacement from the ground -- on the ledge, 55m off the ground.
(ii.) when the ball travels after (i.) and reaches the ground

(i.)
ALWAYS do SUVAT

s = displacement
u = original velocity
v = final velocity
a = acceleration ===>**IMPORTANT -- ensure you know if it's positive or negative due to it's direction.
t = time taken

it's easier to work out the travel of the ball from first release to when it reaches it's maximum height and then returns to the ground [half of part (i.)]:

s = [not given]
u = 31 m/s
v = 0
a = -9.81 m/s^2
t = [not given]

to work out the time [t], you need the 1st equation of motion:

v = u + at
0 = 31 + (-9.81)t
9.81t = 31
t = 31 / 9.81
t = 3.16 s

therefore the TOTAL time in part (i.) is double:

3.16 x 2 = 6.32s

...

part (ii.):

s = 55m
u = 31 m/s
v = [not given]
a = 9.81 m/s^2
t = [not given]

using the 2nd equation of motion to equate v:

v^2 = u^2 + 2as
v^2 = 31^2 + 2(9.81)(55)
v^2 = 2040.1
v = 45.17 m/s

to work out t:

v = u + at
45.17 = 31 + 9.81t
14.17 = 9.81t
t = 1.44 s

...

THUS, the TOTAL time = time of (i.) + time of (ii.)

T = 6.32 + 1.44
T = 7.76
T = 8 [1 s.f.]

Answer 4

6) 8
7) 96
8) 620.7
9) 37
10) 31

The formula I used was scrolling down and looking at the answers... it's called using your resources. Since you have the answers, try working backwards!

Answer 5

I agree with you. Jeff Hardy is entertaining and loves his fans and does lots of stuff for us. You mentioned the 30 foot flip off of a ladder, that is typical Jeff. Jeff is one of my favorites and you never quite know what he is going to do next. You can be sure it will have you on the edge of your seat cheering... Tonight at No Way Out he did a Swanton Bomb off the top of the Elimination Capsule, it was amazing.

Answer 6

formulas: v=u+at, v^2 =u^2+2as , s=ut+0.5at^2, v= final velocity
at max height v=o, u=initial "
s=displacement
a= acc. due to gravity=10m/s^2
or 9.81m/s^2

Answer 7

http://www.collegeboard.com/prod_downloads/ap/students/physics/info_equation_tables_2002.pdf

There are some equations on the second page that may help you but you need to learn this on your own.

Answer 8

you don't understand this becasue you rely on others to understand it for you. do you understand the fundamental flaw in this apporach?

Answer 9

oh, god...do your own homework!