FDownes Q 12. (Grove E6.11 Q13) A gun is aimed at a target on the ground 150m away. If the initial velocity is 125ms-1, find the angles at which the gun could be fired to reach the target (use g = 10ms-2)
Vertical Components
ÿ=-10
ẏ=∫ -10 dt= -10t + C
when t=0, ẏ=125 sin θ
C=125 sin θ
ẏ=-10t + 125 sin θ
y=∫-10t + 125 sin θ dt
y=-5t^2 + 125 t sin θ + C
t=0 , y=0
C=0
y=-5t^2+125 t sin θ
Horizontal Components
ẍ=0
ẋ=∫ 0 dt= 0t +C
when t =0, ẋ=125 cos θ
C=125 cos θ
ẋ=125 cos θ
x=∫ 125 cos θ dt
x=125 t cos θ +C
t=0, x=0
C=0
x=125 t cos θ
We now know our vertical and horizontal motion equations.
We let x=150 and y=0
a) 150=125 t cos θ
b) 0=-5t^2+125 t sin θ
Rearranging a) to get in terms of 't'
t=150/(125 cos θ)
0=-5[(150/125 cos θ)]^2 + 125 [150/(125 cos θ)] sin θ
This simplifies to
0=-5[1.44/(cos θ)^2]+150 tan θ
0=-7.2[1+(tan θ)^2]+150 tan θ
0=-7.2-7.2 (tan θ)^2+150 tan θ
Rearranging,
7.2 (tan θ)^2-150 tan θ+7.2=0
Using the Quadratics roots formula
tan θ=[-(-150) ± sqrt[22500-4(7.2)(7.2)]]/ 2 (7.2)
tan θ= (150 ±149.3072)/14.4
θ=87.25 degrees or 2.75 degrees
OR
θ=87°15' or 2°45' (converted into degrees and minutes as required in the solutions)