1. ## Projectiles question

Ball is projected in a 30m/s velocity with a angle of X from a place called 'O' horizontally.The ball was caught by someone at 1m of height .This is position 'P'.If tan X=4/3 ,what is the time took for the ball to go to 'P' from 'O' ?what is the displacement of the ball at the 'P' position?

I never got the book's answer.here is how I have done it incorrectly :
Y component of 30m/s is 24m/s,X is 18(took from tan x=4/3)

to find the time \uparrow OP s=ut+1/2at²
1=24t+1/2(-10)t²
5t²=24t
t=24/5 <<which is wrong

finding OP distance \to s=ut
s=18*24/5
s=86.5 <<which is wrong

what are the correct ones?
0.447s,1.28m are the answers give in the book.
thanks

2. Hello, silvercats!

Please give us the original wording of the problem.
As stated, the problem is incomplete.
I tried to reword the problem, but it didn't help.

$\text{A ball is thrown with a velocity of 30m/s at an angle }\theta$
$\text{ with the horizontal from a point }O.$

$\text{The ball was caught at a height of 1 m above the ground at point }P.$

$\text{If }\tan\theta =\tfrac{4}{3}\text{, what is the time took for the ball to go from }O\text{ to }P?$
$\text{What is the displacement of the ball at }P?$

$\text{Answers in book: }\:0.447\text{ sec, }\:1.28\text{ m.}$ .??

We need the height of point $O$.
. . Is it on the ground? . . . If so, the answers don't fit.

In that time, its horizontal displacement would be about 9 meters.

3. Originally Posted by silvercats
Ball is projected in a 30m/s velocity with a angle of X from a place called 'O' horizontally.The ball was caught by someone at 1m of height .This is position 'P'.If tan X=4/3 ,what is the time took for the ball to go to 'P' from 'O' ?what is the displacement of the ball at the 'P' position?

I never got the book's answer.here is how I have done it incorrectly :
Y component of 30m/s is 24m/s,X is 18(took from tan x=4/3)

to find the time \uparrow OP s=ut+1/2at²
1=24t+1/2(-10)t²
5t²=24t
t=24/5 <<which is wrong

finding OP distance \to s=ut
s=18*24/5
s=86.5 <<which is wrong

what are the correct ones?
0.447s,1.28m are the answers give in the book.
thanks
That time simply cannot be right unless the thrower is firing the ball upward at the catcher. Possible, but extremely unlikely. (And that doesn't give you the predicted time anyway. It's off by a factor of 10.) My suspicion is that this is the given answer for a different problem.

-Dan

4. I think I have figured out what is wrong.we have to take both x and y parts and combine them using the Pythagorean theorem to get the correct one.I was finding only one direction.thanks for replying

5. Hello, silvercats!

This doesn't help at all . . . it's even more confusing.

I think I have figured out what is wrong.
We have to take both x and y parts and combine them
using the Pythagorean theorem to get the correct one.
I was finding only one direction.

A projectile problem does NOT involve Pythagorus!

$\begin{array}{ccccc}\text{The horizontal displacement is:} & x \;=\;(v_o\cos\theta)t \\ \text{The vertical displacement is:} & y \;=\; h_o + (v_o\sin\theta)t- 4.9t^2 \end{array}$

. . $\text{where: }\;\begin{Bmatrix} v_o &=& \text{initial velocity} \\ h_i &=& \text{initial height} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}$

We are given: . $v_o \:=\:30$

We are given: . $\tan\theta = \tfrac{4}{3}\quad\Rightarrow\quad \sin\theta = \tfrac{4}{5},\;\cos\theta = \tfrac{3}{5}$

Our equations are: . $\begin{Bmatrix}x \;=\; 18t \\ y \;=\; h_o + 24t - 4.9t^2 \end{Bmatrix}$

I'll ask again . . . Please give us the original problem.
. . Otherwise, you're just wasting our time.

6. Thanks for helping.I have found the answer

7. actually ,the question is like that because I translated it to English .that is why it is not that much clear.sorry for that