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I need some help with the following question:

A smooth spherical object (the first object) is projected horizontally from a point of vertical height H = 26.38 metres above horizontal ground with a launch speed (initial velocity) of u = 21.05m/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 $g = 9.81m/s^2$.

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closed as too localized by John Rennie, David Z Mar 22 '12 at 23:31

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Your question is off topic, you don't show us what you have tried, you have a zero percent acceptance rate... –  Raskolnikov Mar 22 '12 at 10:13
    
What @Raskolnikov said. Try this forum instead. –  0sh Mar 22 '12 at 10:19
    
@Raskolnikov: Thanks for the input –  Xavier Mar 22 '12 at 14:33
    
Yeah, Physics Forums is a better place for asking for help on a specific homework problem (though they also expect you to show work there). –  David Z Mar 22 '12 at 23:32
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1 Answer 1

Split the problem into horizontal and vertical components:

For the same range and flight time, the horizontal components of two objects' velocities must be equal (so 21.05 m/s).

The flight time of the first object is just the time it takes a stationary body to fall from a height of 26.38 metres -- can you work that out? So now calculate the initial vertical speed to give the second object so that it falls back to earth after the same flight time.

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The flight time of the first object = 1/2gt^2 = 26.38, which eventually gives 2.32 secs. Would the horizontal range be 2.32*21.05 = 48.836m? or should I take into account of the 1/2at^2 portion as well? –  Xavier Mar 22 '12 at 14:31
    
You don't need to know the horizontal range. You just need the horizontal and vertical components of the initial velocity. –  TonyK Mar 22 '12 at 16:09
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