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A smooth spherical object (the first object) is projected horizontally from a vertical height of $26.83$ metres above horizontal ground with a launch speed of $23.44\textrm{ ms}^{-1}$.

A second identical object is projected from ground level with launch speed, $V$ and at an angle above the horizontal. Calculate the launch angle in radians, if the second object is to have the same horizontal range and same time of flight as the first object. Given $g=9.81$.

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You should be able to get the flight time and distance of the first object easily using basic motion equations. Is the launch angle supposed to be in terms of V? There is a simple equation for the trajectory of an object which should get you to the answer. – chris Mar 26 '12 at 3:15
The same question was asked here and it was migrated to physics.SE here! There is an answer there while it is closed as too Localized. I do not what is to be done this time! – user21436 Mar 26 '12 at 5:20
up vote 3 down vote accepted

You probably know that (for reasonable $t$) the vertical distance travelled by the first object after $t$ seconds is $\frac{1}{2}at^2$, where $a$ is the acceleration. So if $t_1$ is the time until the first object hits the ground, we have $$26.83=\frac{1}{2}(9.81)t_1^2.$$ Now we can compute $t_1$.

Look at the second object. Since time of flight and horizontal distance travelled is the same as for the first object, the horizontal speed $v_1$ of the second object is $23.44$.

Let the initial vertical speed be $v_2$. Then after time $t_1/2$, the second object reached maximum height. The vertical velocity of the second object, after time $t$, where $t\le t_1$, is equal to $v_2-9.81t$. At time $t_1/2$, the vertical speed reaches $0$, and therefore $$v_2-9.81t_1/2=0.$$ Since we know $t_1$, we now know $v_2$.

The launch angle $\theta$ is easy to compute once we know $v_1$ and $v_2$, because $$\tan\theta=\frac{v_2}{v_1}.$$

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