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#1
Dan, or anyone else, got a prob on a few questions and hoping for your assistance,

Jan 2000, old style M2 question 9

q In a serve at tennis, a tennis ball is given an intial speed of u ms-1. Ball is struck at P, 2.8 m directly above O. Ball passes over net vertically above ground at pt N 0.9m high and where ON=12m

Assuming the ball is initially struck horizontally, Fing to 1 dp the least value of u which will enable the ball to pass over the top of net.

Thanks

Also review exercise 1 M2 Question 9, im stuck on.
0
16 years ago
#2
right, this is how I do it:

I hope you understand the x x. x.. notation. (where x means the displacement of x, x. means the velocity of x, x.. is the accleration of x)

From what you say:

x = ut
x. = u
x.. = 0

y = 2.8 - 0.5gt^2
y. = -gt
y.. = -g

Thus, you know that ON = 12m

That means that the displacement of X is 12m.

So ut = 12

you also know that it just clears a net 0.9m above point N.

so y = 0.9

2.8 - 0.5gt^2 = 0.9
1.9 = 4.9t^2
t = 0.62 seconds (reject the minus answer)

since ut = 12
u = 12/t = 12/0.62 = 19.27ms-1

I hope thats right anyways!
0
16 years ago
#3
Done it differently:

Use suvat equeations horizontally and vertically:

Vert:

u=0
a=-9.8
s=-1.9
t=T

Horiz:

u=u
v=u
a=o
s=12
t=T

From vertical

s=ut + 0.5 at^2
-1.9 = 0 - 4.9t^2

t^2 = 1.9/4.9

square root take positive:

t= 0.62 seconds as byb3 states. So same value of u:

Hope this helps if you did not understand differetiation of vectors method: I will post a solution to the other question from the Review Ex when I get home, all the best in the exam. Dan
0
16 years ago
#4
Just done this, nice question!!!

Vertically (1)

U = 49Sinx
a = -9.8 m/s/s
s = -3 4/15
t = T

Horizontally (2)

U = 49Cosx
a = 0
s = 98
t = T

Using s=ut+0.5t^2 on (2)

98 = 49Tcosx

T = 2/cosx
T = 2Secx

Now same formula on (1) using this value of T

S = ut + 0.5at^2

-3 4/15 = (49sinx)(2secx) + 0.5 (-9.8)(2secx)^2

multiply by 15:

-49 = 1470tanx - 294(sec^2 x)

Now sec^x = 1 + tan^x

Sub in

-49 = 1470 tanx -294(1 + tan^2x)

Divide by 49:

-1 = 30 tanx - 6(1+tan^2 x)

Expand and collect on one side
6tan^2 x -30tanx + 5 =0 QED

b) Quadratic formula on the above

Gives tanx = (30 +- root 780)/12

So x = 10, 78 degrees on for each value of tanx

c) Work out T for each value of x

NB T= 2/cosx

For x =10 T = 2/cos10 = 2 secs
x= 78 T = 2/cos78 = 9.6 ses

Answer = 2 seconds to nearest sec (shortest time)

Hope this helps dude. All the best, DAN
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