A smooth spherical object (the first object) is projected horizontally from a po

Question

A smooth spherical object (the first object) is projected horizontally from a point a vertical height H = 25.76 metres above horizontal ground with a launch speed u = 24.39 ms-1.
A second identical object is projected from ground level with launch speed v and at an angle ? above the horizontal. Calculate the value of this launch angle ? 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.

Explanation / Answer

time of flight = sqrt(2H/g) = sqrt(25.76*2/9.81) = 2.2916 m/s

horizontal range = vt =24.39*2.2916 = 55.892 m

now for second object

vertical speed = V*sin p where p= angle of launch

we know that for vertical motion final velocity is zero

and upward time = t/2 = time of flight /2

so Vf= Vi-at/2=> V sin p - at/2 =0

so V sin p = 9.81*2.2916 = 22.4806/2 = 11.2403

now horizontal speed is

v cos p

so x = V cos p *t + 0.5*g*t^2

solve we get

V cos p = 55.892/2.2916 - 0.5*9.81*2.2916 = 24.3899-11.24029 = 13.1496

now

V sin p / V cos p = 22.4806/13.1496 = 1.7096/2

now tan p = 1.7096/2 = 0.854801

=> p =arctan(1.7096/2) = 0.707 radians

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