I will skip the derivation and cut to the final equation for horizontal range X:
X = (V^2/g)sin 2A
where A = the launch angle.
24 = (400/9.8) sin 2A
sin 2A = 0.588
2A = 36 or 144 degrees
A = 18 or 72 degrees
Looks like you did it right. Consider the possibility of two solutions.
An object is shot at 20m/s at an angle theta. The object lands 24m away. What is the angle?
Can you check if I am doing this right?
I had
v(initial) = 20 m/s
g = 9.8 m/s^2
x = 24m
1st I solved for time:
x = v(initial) t
x = v(initial) cos(theta) t
24 = 20 cos(theta) t
24 / [20 cos(theta)] = t
Y-motion
y = v(initial) sin(theta)*t - (1/2)gt^2
0 = [20sin(theta)]*[24 / 20cos(theta)] t -(1/2)(9.8)[24 / 20cos(theta)]^2
0 = 24tan(theta) - (7.056/cos^2(theta))
7.056 = 24tan(theta)cos^2(theta)
7.056 = 24sin(theta)cos(theta)
<then I divided both sides by 12 to get a trig identity>
.588 = 2sin(theta)cos(theta)
.588 = sin2(theta)
theta = arcsin(.588) / 2
theta = 18.008 <--answer
Is that how you would do it? Is that the correct answer?
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