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.
Vh1= Horiz. Veloc. Proj. 1 (on t=0).
Vv1= Vert. Veloc. Proj. 1 (on t=0)
Vh2= Horiz. Veloc. Proj. 2 (on t=0).
Vv2= Vert. Veloc. Proj. 2 (on t=0)
t=time ti= time on impact
q=requested angle
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Projectile 1: Vv1/Vh1=tg 30°
Projectile 2: Vv2/Vh2=tg 60°
Projectile 1:
on time t Vv(t)=Vv1-gt
on time ti Vv(ti)= -Vv1 = Vv1 - gti
=> ti = 2Vv1/g = (2Vh1*tg 30°)/g (1)
Projectile 2:
distance a from P to pole
a = Vh2 *ti = (2Vh2*Vh1*tg 30°)/g
But as Vh2=Vh1 (same t=0 and t=ti ) (2) =>
a=2*(Vh1)^2*tg 30°/g
vertical elevation of proj. 2
h= Vv2*t - g(t) ^2/2
on t=ti h= Vv2*ti - g(ti) ^2/2
From (1) and considering also (2) and (3):
h = (Vh1 )^2)/3g
Then:
h/a= [(Vh1 )^2/3g]/[2*(Vh1)^2*tg 30°/g]
= 1/3(2*tg 30°)= tg 60° - tg 30°
q= arctg h/a = 16.10°
×++
error arctg h/a=49.10°
Edited on March 24, 2018, 3:05 pm
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Posted by armando
on 2018-03-24 09:33:47 |
spoiler of precedent post
