0
$\begingroup$

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 $A$

  • 2
    Dear @sfclpsml. Welcome to math.SE. You might find a lot of help here if you follow the guidelines under the [FAQ section](http://math.stackexchange.com/faq). 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.2012-03-25

1 Answers 1