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A person wants to throw an object from the top of a tower $9,0m$ high towards a target which is $3,5m$ far from the place where the person is launching the object. Suppose that this object is thrown horizontally.

What I want to find out is:

  1. What initial speed does the object need to have in order to hit the target?
  2. What is the acceleration of the object one moment before it touches the ground?

I suppose that the answer to the second question is simply $9,81m/s^2$, but I am not sure about it because it looks so simple :D

As far as the first question is concerned, I was thinking about using the formula $y_{MAX} = \frac{v_0^2 + sin^2\theta}{2g}$, with $y_{MAX}$ being $3,5m$ and $\theta$ being $45°$, but the result in this case would be extremely unrealistic ^^

Am I going in the right direction (in both questions)?

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closed as off topic by dmckee Jun 1 '12 at 18:27

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

    
As far as (1) and using the $y_{max}$ formula, that formula most likely was derived in your notes or textbook somewhere. I would suggest looking back at the assumptions that led to the final form and see if they deviate from the problem at hand. For (2) have you thought about how to calculate the acceleration on an object in general? That is, many students see 'acceleration = blank' and throw in g because it seems reasonable. It might be correct in this case, but I would familiarize yourself with the MOST GENERAL METHOD to get acceleration, and see if it indeed gives you g in this case. –  DJBunk Jun 1 '12 at 17:14
    
@DJBunk thanks for your reply. I will have a look at (1) then. For (2), isn't acceleration given by $v/t$? But I don't have $v$ nor $t$. –  user1301428 Jun 1 '12 at 17:35
    
Welcome to Physics.SE. Our site policy is that we will answer very basic conceptual questions (say about the rules of projectile motion) but will not answer basic home work questions. As phrase this is homework question. To understand the difference note that kηives provided you with a general, conceptual answer. You can get the question re-opened by editing to a general form. –  dmckee Jun 1 '12 at 18:23

1 Answer 1

up vote 1 down vote accepted

You are right about the second question.

for the first one. Ask yourself how long (time) does it have before it hits the ground (and what is this dependent on)? Some thought will reveal that the time it's in the air has to do only with its motion in the $y$ direction. So, if we only throw the ball horizontally, then the equation for how long with will be in the air comes from $$y=\frac{1}{2}gt^2\rightarrow t=\sqrt{\frac{2y}{g}}$$ That is how long the object has before it comes in contact with the ground. Now, we also know how far we want the ball to get in the horizontal direction, 3.5 m. What equation has no acceleration in the $x$, and relates velocity, distance, and time? $$x=v_xt\rightarrow v_x=\frac{x}{t}=x\sqrt{\frac{g}{2y}}=\sqrt{\frac{x^2 g}{2y}}$$

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Thank you very much for your answer. I have one question though: if the object follows a projectile motion, why are we never using one of the formulas related to the projectile motion here? –  user1301428 Jun 1 '12 at 18:03
    
My go-to projectile motion equations are $y=\frac{1}{2}at^2+v_0 t+y_0$, $v_f=at+v_0$, $v_{f}^2 -v_{0}^2=2a\Delta x$. They can do almost anything if used correctly and amongst themselves, and are good in any direction. –  kηives Jun 1 '12 at 18:08
    
While this is a nice conceptual answer completely in keeping with our policy on basic homework questions, my personal (i.e. not ♦-powerd) opinion is that by providing it before the OP has been required to put his questions into general terms, you are effectively by-passing the rule and harming the site. –  dmckee Jun 1 '12 at 18:21
    
my bad, I'll keep that in mind in the future and won't do it again. –  kηives Jun 1 '12 at 18:57

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