projectile

[tony873004]

A projectile fired from ground level at 25 m/s hits the ground 31.25 m away. At what angle was it fired?

I wrote equations for t for both x and y

t = delta x / velocity x
t = 2 * viy/a

and set them equal to each other to eliminate t
and replaced velocityx with cos(theta)*25
and replaced velicityinitialy with sin(theta)*25

delta x / cos(theta)*25 = 2*sin(theta)*25/a

25 / cos(theta)*25 = 2*sin(theta)*25/9.81

sin(theta) * 25 * cos(theta)*25 = 153.28

My trig is not that good. How do I continue? Or is this even the right way to tackle this proble? Is there an easier way?

[Muzza]

Assuming your previous work is correct, use the fact that sin(t)cos(t) = sin(2t)/2 to solve the last equation.

[tony873004]

Assuming your previous work is correct, use the fact that sin(t)cos(t) = sin(2t)/2 to solve the last equation.
Thank you! That was exactly the trigonometry trickery I was looking for. :rofl:

My problem now boils down to
\frac{sin(2theta)25^2}{2} = 153.21825

2theta = sin^{-1} = 0.4905

2theta = 29.37

theta = 14.685

The back of the book, which rounds for significant figures list 15 degrees and 75 degrees. So I got one of the two answers with this method. But sin (75*2) also equal 0.5. And it makes sense that both the 15 and 75 degree angles should yield the same result in this type of projectile problem.

So how do I make sin^{-1} acknowledge that sin^{-1}0.5 equals both 30 (15*2) and 150 (75*2)?

[Leong]

for your case, 0 \leq \theta \leq 90\ degree, so
0 \leq 2\theta \leq 180\ degree
so, you have to find all the values of theta that are in the above range and you have 2 values : 30 degree and 150 degree. so theta equals to 15 degree or 75 degree.