- #1
devanlevin
a tank shoots at a target on a mountain 10km away from it, the hight of the mountain is 500m the tank shoots at a speed of 1800km/h, at which angles can the tank aim?
this is a question in kinematics and the only force to take into account is gravity,
using
V(t)=Vo+at
X(t)=Xo+Vot+½at²
V²-Vo²=2aΔx
Vx=const=cosθ*Vo
Vy(t)=sinθ*Vo+at
Vo=500m/s
Xf=10000m
Yf=500m
knowing that the target is at (10000,500) i use the equation for X,
X(tf)=500cosθ*tf=10000m
tf=20/cosθ
Y(tf)=500m=500sinθ*(20/cosθ)-4.9*(20/cosθ)²
=1000tanθ-1960/(cosθ)²
now from here i don't know what to do, one equation with both tanθ and cosθ, how do i bring them to some common expression?? i have had a very similar quesion before which i too could not get past this point.
since i do not have T, i thought of using the 3rrd equation, but i don't see how i can
this is a question in kinematics and the only force to take into account is gravity,
using
V(t)=Vo+at
X(t)=Xo+Vot+½at²
V²-Vo²=2aΔx
Vx=const=cosθ*Vo
Vy(t)=sinθ*Vo+at
Vo=500m/s
Xf=10000m
Yf=500m
knowing that the target is at (10000,500) i use the equation for X,
X(tf)=500cosθ*tf=10000m
tf=20/cosθ
Y(tf)=500m=500sinθ*(20/cosθ)-4.9*(20/cosθ)²
=1000tanθ-1960/(cosθ)²
now from here i don't know what to do, one equation with both tanθ and cosθ, how do i bring them to some common expression?? i have had a very similar quesion before which i too could not get past this point.
since i do not have T, i thought of using the 3rrd equation, but i don't see how i can