0

A golf ball is hit in a horizontal direction off the top edge of a building that is $100$ feet tall. How fast must the ball be launched to land 450 feet away?

I have some of the equations that I have been working with here:

$$ \begin{align*} \mathbf{s}(t) &= \langle v_0t \cos(\theta),v_0 t \sin(\theta) - \tfrac{1}{2}g t^2 \rangle \tag{1} \\ \mathbf{v}(t) &= \mathbf{s}^{\prime} (t) = \langle v_0 \cos(\theta),v_0 \sin(\theta) -gt \rangle \tag{2}\\ \mathopen|\mathbf{s}^{\prime}(t)\mathclose| &=\sqrt{v_0^2-gt} \tag{3} \end{align*}$$

where $(1)$ is the position of the object at time $t$, $(2)$ is the velocity of the object at time $t$, and $(3)$ is the speed of the object at time $t$ and $g$ is the gravitation constant equaling $32$ feet per second squared.

With the given information I'm not sure how to proceed with the problem. Ideally I would want to start use $(3)$, but I don't know the value of $v_0$. I know that I want the object to travel a horizontal distance of $450$ feet, so the horizontal component of $\mathbf{s}$ is equal to $450$, however, again, I confound my problem with unknown values for $v_0$ and $\theta$.

Any hints or guidance for this seemingly trivial problem would be greatly appreciated.

Alana
  • 179
  • I'm guessing it's not around other buildings ? –  Sep 24 '17 at 01:41
  • Yeah, it is not around other buildings. – Alana Sep 24 '17 at 01:42
  • Only one problem: you're mixing up feet and metres by using 9.8 instead of 32.2 as the standard gravity. – Parcly Taxel Sep 24 '17 at 01:46
  • calculate the drop time to the ground with the distance under constant acceleration formula that will give you a time you might have to get it the 450 feet it needs if you don't want to roll on from there( negating air resistance) . –  Sep 24 '17 at 01:46

1 Answers1

0

Take $g = 32 \,\text{feet/sec}^2$. $S = \frac{1}{2}gt^2$. Thus, it takes time $\sqrt{\frac{2S}{g}} = \sqrt{\frac{200}{g}} = 2.5\,\text{secs}$ for the ball to reach the ground. Now, in 2.5 seconds, the ball must travel 450 feet. Thus, the speed must be $\frac{450}{2.5} = 180 \,\text{feet/s}$.

EDIT: The original poster asked for a visualization of what I did, hence this edit.

enter image description here

Forgive me for the crudely drawn diagram, but this is what I visualized. Since the ball is hit horizontally, the velocity component is completely along the horizontal direction. There is no component along the vertical direction.

Now, the matter of whether to use + or - with the $\frac{1}{2}at^2$ term is just a matter of where you are looking from. If you are standing at the bottom and throwing a ball up in the air, then you have to use the negative sign.

So, given this visualization, what you have to realize is that since the velocity is $u$ feet per second along the horizontal direction, in time $t$ seconds, the ball will travel a total of $ut$ feet. Thus is because I used $S = ut + \frac{1}{2}at^2$. Note that the acceleration along the horizontal direction is zero (the problem could be more complicated if someone says that the ball was hit with some $x$ Newton amount of force. Then you would calculate acceleration appropriately), so $S = ut$. Since you know $S$, and what you want to calculate is $u$, you just need $t$, i.e. the time for which the ball travels horizontally to complete your calculation.

The time for which the ball travels horizontally is obviously determined by how long the ball takes to reach the ground. We know that the ball has 100 feet to travel in the vertical direction and we know that there is no initial velocity along the vertical direction. Thus, the only thing that is making it move vertically is the gravitational acceleration. That is why I used $+\frac{1}{2}gt^2$ to determine how much time it will take to reach the ground.

Hope this helps.

Abhiram Natarajan
  • 1,226
  • depends on what angle it's shot at as well. also at g=32 feet/second you have under 3 seconds until it hits the ground based on your math. –  Sep 24 '17 at 01:53
  • The question says "hit in horizontal direction". The time you have is exactly 2.5 seconds, i.e. $\sqrt{\frac{100}{16}}$. – Abhiram Natarajan Sep 24 '17 at 01:56
  • then where do you get the 20 seconds part ? –  Sep 24 '17 at 01:57
  • Where is the 20 seconds? – Abhiram Natarajan Sep 24 '17 at 01:57
  • third sentence in your answer. –  Sep 24 '17 at 01:58
  • OOPS! Horribly typo. It should just be $\sqrt{\frac{200}{g}}$. Will edit right away. THANKS! – Abhiram Natarajan Sep 24 '17 at 01:59
  • Sorry, I deleted my last comment, because I realized that $v_0 = 0$, because the object is at rest, so I only have one question: Why do you have $\tfrac{1}{2}gt^2$ instead of mine which has a negative sign? I'm trying to visualize what's going on, so could you also explain what, in your mind, traveling at a horizontal distance on a building $100$ feet tall even means? I just want to make sure I'm visualizing the same thing as you. – Alana Sep 24 '17 at 02:08
  • Edited the answer. Hope that helps. – Abhiram Natarajan Sep 24 '17 at 02:36