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# projectile motion

ivan_07
a ball is thrown at an initial velocity of 30m/s at an angle θ with the horizontal. it hits point B, 30 m. below the point of origin in exactly 4 sec.

compute the value of θ

x=V(cosθ)t
x=30cosθ(4)
x=120cosθ

y=x(tanθ) - gx^2/2v^2cos^2θ
-30=120(Cosθsinθ)/cosθ - 9.81(120cosθ)^2/2(30)^2cos^2θ
9.81(120)^2/2(30)^2 -30 =120 sinθ
48.48=120 sinθ
θ=23.80

am i right with my answer???
Bikerman
 ivan_07 wrote: a ball is thrown at an initial velocity of 30m/s at an angle θ with the horizontal. it hits point B, 30 m. below the point of origin in exactly 4 sec. compute the value of θ x=V(cosθ)t x=30cosθ(4) x=120cosθ
Yep that's correct
 Quote: y=x(tanθ) - gx^2/2v^2cos^2θ

I get y=x*tanθ - gx^2/2V^2(1+tan^2θ)
I don't think cos^2θ = (1+tan^2θ) - though I haven't done trig identities for a coon's age - so that means one of us has it wrong....
Probably me, but worth having another look at that step perhaps:
I get there via substitution to give:
y = V.sinθ.x/V.cosθ-g/2(x/V.cosθ)^2
Do we agree to that point?
chatrack
Hi,
You are correct

considering vertical motion only,
u= usin8
S= -30m
a=-g=-10m/s2
t=4sec

-30 = usin8 -1/2 x 9.8 4x4
jajarvin
 ivan_07 wrote: a ball is thrown at an initial velocity of 30m/s at an angle θ with the horizontal. it hits point B, 30 m. below the point of origin in exactly 4 sec. compute the value of θ x=V(cosθ)t x=30cosθ(4) x=120cosθ y=x(tanθ) - gx^2/2v^2cos^2θ -30=120(Cosθsinθ)/cosθ - 9.81(120cosθ)^2/2(30)^2cos^2θ 9.81(120)^2/2(30)^2 -30 =120 sinθ 48.48=120 sinθ θ=23.80 am i right with my answer???

I get this:

x(t) = V*cos(θ)*t
now
t = 4 seconds
and
x(t=4) =30m
V = 30m/s
so
30 = 30 *cos(θ)*4
or
*cos(θ) = 1/4
or
θ = 75.52