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A smooth spherical object (the first object) is projected horizontally from a point of vertical height $H = 28.71$ metres above horizontal ground with a launch speed (initial velocity) of $u = 22.68 \operatorname{m/s}$

A second identical object is projected from ground level with launch speed (initial velocity) $v$ and at an angle $A$ above the horizontal.

Calculate the value of this launch angle $A$ in radians, if the second object is to have the same horizontal range and the same time of flight as the first object. Take acceleration at gravity to be $g = 9.81\operatorname{m/s}^2$

Attempt answer

$(1)$ Use $t = \sqrt{\dfrac{2H}g}$ to find $t$

$(2)$ $t = \dfrac{2u\sin A}g$ to find $A$

sfclpsml
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    Dear @sfclpsml. Welcome to math.SE. You might find a lot of help here if you follow the guidelines under the FAQ section. For this question, would you please include, in addition to the question, the effort you did? Also, please indicate in which part you got stuck. This will help other users give you useful and pedagogical hints. –  Mar 25 '12 at 05:33

2 Answers2

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Hints:

(1) Since the initial vertical component of velocity is $0$ your first formula will give you the time when the object hits the ground. You have need to calculate how far away that is using $u$ and $t$.

(2) In your formula, your formula uses $u$ when it should be using $v$. If you replace it then your formula based on the vertical component of velocity would be true, but would have two unknowns $v$ and $A$. You also need a second formula involving $v$ and $A$ looking at the horizontal component of velocity and distance, using the result you should have calculated from (1). You then need to either eliminate $v$ to find $A$ or to solve for $v$ and then use that to solve for $A$; the former should involve taking an arctangent.

Henry
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Unfortunately, this problem has an ambiguity in it; it is not clear whether it was intentional or not. The mention of the "smooth spherical object" would seem to be the problem-poser's way of indicating that air resistance may be neglected. Under that assumption (and the "flat ground" assumption), the range of the projectile would be given by $ \ R \ = \ u·T \ , $ with $ \ T \ $ being the time from launch to impact. With the information given and the formula you have for the time the projectile will take to fall, we can calculate $$ T \ = \ \sqrt{\frac{2·H}{g}} \ \approx \ 2.42 \ \text{seconds and} \ \ R \ \approx \ 54.9 \ \text{meters .} $$

The difficulty arises with the second projectile: if it is to attain the same horizontal range in exactly the same amount of time, then (in the absence of air resistance) it only needs to have the same horizontal speed as the first one. The horizontal component of the launch velocity is $ \ v_h \ = \ v \cos (A) \ \ , $ so we have $$ R \ = \ u·T \ = \ v \cos (A) · T \ \ \Rightarrow \ \ \ v \cos(A) \ = \ 22.68 \ \text{m/sec} \ \ .$$

Since there is no further specification for either the launch speed or angle, there is not a unique solution to this problem.

[I was amused by the stipulation of a "smooth spherical object". Smooth spheres have notoriously poor flight characteristics because of turbulent flow around them. This is why firearms and artillery switched from musket- and cannon-balls to bullets and shells and why golfballs have (carefully designed) dimples.]