There is a vertical pole perpendicular to the horizontal plane. From point P on the plane, 2 projectiles are fired simultaneously at different velocities. The first projectile is fired at an angle of 30o and it hits the foot of the pole. The second projectile is fired at an angle of 60o and it hits the top of the pole
It is further known that the projectiles hit the pole at the same time.
Find the angle subtended by the pole from P.
The trajectory of a projectile:
x = v*t*cos(w)
y = v*t*sin(w) - g*t^2/2
where g = acceleration of gravity
v = velocity at launch
t = time after launch
w = angle at launch
Let a = distance from launch point to the bottom of the pole
h = height of the pole
Equations of projectile hitting bottom of the pole:
a = v_3*t*cos(30)
= v_3*t*sqrt(3)/2 (1)
0 = v_3*t*sin(30) - g*t^2/2
= v_3*t/2 - g*t^2/2 (2)
Equations of projectile hitting top of pole:
a = v_6*t*cos(60)
= v_6*t*/2 (3)
0 = v_6*t*sin(60) - g*t^2/2
h = v_6*t*sqrt(3)/2 - g*t^2/2 (4)
Combining (1) and (3) we get
v_6 = v_3*sqrt(3) (5)
Combining (2) and (4) we get
h = v_6*t*sqrt(3)/2 - v_3*t/2 (6)
Combining (5) and (6) we get
h = v_3*t (7)
Combining (1) and (7) we get
h/a = 2/sqrt(3)
Therefore, the angle subtended is
arctan(h/a) or approximately 49.1066 degrees.
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Posted by Bractals
on 2018-03-24 13:24:19 |
Solution
