Question

A tennis ball is served horizontally from 2.4m above the ground at 30m/s
a) the net is 12m away and 0.9m high.. Will be ball clear the net?
b) where will the ball land.. ?


Can someone please explain how to handle these conditions ( in which two heights are given)..

Answer 1

Ok, what you want to know is: will the ball fall the distance of 2.4m - 0.9m = 1.5 m by the time it reaches the vertical plane of the net. Make sense?

Answer 2

yeah..

Answer 3

Ok. As gravity is the only force acting on the ball after it is served, how far does it fall in time t? In other words, let's first calculate how long it would take for the ball to fall that far.

Can you do that?

Answer 4

yeah I can

Answer 5

ok ... calculate it and tell me what you get.

Answer 6

sorry.. can tell me which equation should i use for it.. ?

Answer 7

s= vt ?

Answer 8

In time \( t \), the ball falls a distance of
\[ y = \frac{1}{2}gt^2 \]

Answer 9

Therefore, time time required to fall a distance of \( h = 1.5 \) m is
\[ t = \sqrt{\frac{2h}{g}} \]

Answer 10

ok got it, then..?

Answer 11

Now after you've found that time, ask yourself:
How long does it take the ball the reach the vertical plane of the net? Is it before or after the time we just calculated?

If it is before, the ball clears the net.
If it is after, the ball does not clear the net.

Answer 12

Ok, the approach in my words, we need to find the vertical component (y) after 12m in horizontal basically, and see if it crosses the net.

Answer 13

As a world-class tennis player, I'm sure that'll be straight-forward for you. ;-)

Answer 14

:D

Answer 15

I am not world class lol :D

Answer 16

Still trying to solve this trust me, i have 2 books open in front of me lol

Answer 17

haha.. I've just one :)

Answer 18

wtf how to write in that

Answer 19

ok now we are on the court, it more looks like

Answer 20

yeah , definitely :P

Answer 21

Gotta go now, will see your work later, arcticf0x ;)

Answer 22

You mean you haven't found the solution yet? What time did you find for the first part?

Answer 23

its fired from a height H, so range is \[u\sqrt{2H/g}\]

Answer 24

and we need something to do in time to get that .9m into account

Answer 25

As I said, what we want to know is how long it takes the ball to fall to the level of the net. I.e., how long it takes to fall a height of
\[ h = 2.4 - 0.9 = 1.5 \ meters \]

Answer 26

hey can any of you confirm, is the height at net 13.01 m?

Answer 27

Now, because the only force acting on the ball after serving is gravity, and because the ball initially has no vertical velocity, the distance it falls in time t is
\[ y = \frac{1}{2}gt^2 \]
Solving now for \( y = h = 1.5 \ m \), we find \( t = \sqrt{2h/g} = 0.55 \ seconds \).

Answer 28

make it s simple AS possible,in the horizontal direction,we do not have any acceleration so only horizontal speed=30m/s
so speed=distance/time
so time=0.4 seconds
that is the time taken for ball to cover a horizontal range of 12m
so next we have acceleration g in the vertical direction so
but no initial veleocity to begin with as everythiong in horizontal later
u=0
s=1/2at^2
so we have time 0.4
a=g=10
then displacement OR DISTNCE FROM THE GROUND LEVEL(HEIGHT) should be greater than the nets height(so that it is avbove it) that u got by calculation
so now i would be happy if u said whether the ball crosses or not

Answer 29

Now, how long does it take for the ball to travel to the vertical plane of the net? Well, it has an initial velocity in the horizontal direction of v = 30 m/s. As there is no acceleration in the horizontal direction, its velocity is constant at that rate. Now, because
v = (distance)/(time),
we have
time = (distance)/v = 12/30 = ...

Answer 30

... 0.4 seconds as you say.

Hence the ball reaches the vertical plane of the net before it reaches the height of the top of the net. Therefore the ball clears the net.

Answer 31

Wow.

Answer 32

Wish i could have thought of that myself ;P You are a genius :D

Answer 33

Now, Can completely understand ;)

Thanks JamesJ