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I'm an aspiring mathematician and physicist and have been self-teaching myself physics. I've been reading Raymond A. Serways book known as 'Physics for Scientists and Engineers 9th Edition Ohio State University Edition (With Modern Physics)'

A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, forming a 1.00 m high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of 53.0° above the horizontal at a point 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

(a) Find the speed at which the ball was launched.

(b) Find the vertical distance by which the ball clears the wall.

(c) Find the distance from the wall to the point on the roof where the ball

First off, here's how I solved (a)

I used $$d=v \cdot cos\theta \cdot t $$ 2nd I did some basic algebra $$d \cdot v=\frac{cos\theta \cdot t}{d}$$ 3rd I ended up with this equation (which is correct) $$v = \frac{d}{cos\theta \cdot t}$$ which we substitute in our numbers and it becomes $$v=\frac{24.0m}{(cos(53^\circ) \cdot 2.20s)} = 18.1m/s$$

I'd like to know where I move on for (b) and (c).

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closed as off-topic by sammy gerbil, ZeroTheHero, AccidentalFourierTransform, DarenW, David Z Feb 7 '18 at 6:32

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  • $\begingroup$ Whats with the downvotes? If theres anything I need to improve I'm more than willing to. $\endgroup$ – Graham Best Feb 7 '18 at 3:58
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    $\begingroup$ Sorry this is not a problem solving site. Suggestions for alternative sites can be found here. $\endgroup$ – sammy gerbil Feb 7 '18 at 4:00
  • $\begingroup$ stay away from “check my work” questions and concentrate on concepts rather than problems. The question as such is “off-topic” and should be closed rather than downvoted. $\endgroup$ – ZeroTheHero Feb 7 '18 at 4:06
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To solve for b, you can use the initial velocity and angle to create a function of height with respect to time: $ \vec{y} (t) = \vec{v_y} \Delta t + \frac{\vec{a}}{2}\Delta t^2$ . This doesn't immediately help you, but you can make a substitution to change $\Delta t $ into $x$.

$\vec{x} = \vec{v_x}\Delta t$ , so $ \Delta t =\frac{\vec{x}}{\vec{v_x}} $. You know your x velocity, so you can make the substitution to create

$\vec{y} (x) = \frac{\vec{v_y}}{\vec{v_x}} x + \frac{\vec{a}}{2} \frac{x^2}{\vec{v_x}^2}$ With this, you can plug in the x value that corresponds to the distance to the wall and get the height when the ball passes the wall.

Your question got cut off, but I'll try to answer the question as I guess it to be: For part c, you can use x velocity and the time to find horizontal distance. Because you know that the landing height is 6m, you can use the Pythagorean theorem to solve for total distance. If you need any further help, message me, and I'd be happy to complete the problem.

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  • $\begingroup$ Ahah! Thank you so much for your great answer. C seems to be right, I'll give it a go and report back. By the way, sorry for the off-topic question. $\endgroup$ – Graham Best Feb 7 '18 at 4:18

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