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.
(In reply to
spoiler of precedent post by armando)
Upon reading armando's explanation, I realize I neglected the statement that the two projectiles hit the pole at the same time.
That being the case V1*cos(30°)=V2*cos(60°).
V2=V1*cos(30°)/cos(60°)
V2/V1 = tan(60°) ~= 1.732050807568877
From this I get an angle of 49.107°
I see that this differs from armando's 16.10°, which needs further investigation.
I do find peculiar armando's
h/a= [(Vh1 )^2/3g]/[2*(Vh1)^2*tg 30°]
= 3/(2*tg 30°)= tg 60° - tg 30°
q= arctg h/a = 16.10°
First 3/(2*tg 30°) ~= 2.59807621135332
while tg 60° - tg 30° ~= 1.15470053837925
neither of which have an arctg of 16.10°. The former has an arctan of 68.95°, and the latter has an arctan of 49.107°, in agreement with my current answer.
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Posted by Charlie
on 2018-03-24 13:11:48 |
considering the simultaneity issue
