# Projectile Question

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 $g = 9.81\operatorname{m/s}^2$

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

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

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Please don't use [homework] as the only tag for your questions. Find another relevant tag that describes the content, not just what it means to you. Thank you. –  Arturo Magidin Mar 25 '12 at 4:15
What have you tried? –  user5137 Mar 25 '12 at 4:50
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. –  user2468 Mar 25 '12 at 5:33
(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.