Question

A smooth spherical object (the first object) is projected horizontally from a point a vertical height H = 25.02 metres above horizontal ground with a launch speed u = 22.06 ms^-1.
A second identical object is projected from ground level with launch speed v and at an angle θ above the horizontal. Calculate the value of this launch angle θ in radians, if the second object is to have the same horizontal range and the same time of flight as the first object. Give your answer to 3 decimal places and take g = 9.81 ms^-2.

Answer 1

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solution- The hooizontal speed of the frost object is 22.06 mb The height of this object launch can be used to find The time of flight. 6-2258}m/ The sange of this object is de vat = 22.06* 2.258 = 149.82)n Foo the 2nd object the angle canbe taken as the tan inverse of the salio of the post object fallen and the sange of the object. since at has the same sange of the first object O= Tan ( height of the doopped object) sange - Tak ( 23.02) - Tan'(0-5622) To= 26.67) 49.82