Tuesday

March 31, 2015

March 31, 2015

Posted by **Dreskillz89** on Tuesday, September 22, 2009 at 10:02am.

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?

- Physics -
**drwls**, Tuesday, September 22, 2009 at 11:56amI 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.

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