# Projectile motion at an angle

1. Dec 13, 2004

### kreil

A slug is shot at a 50 degree angle from a launcher 1.15m off the ground and has a muzzle velocity of 3.37m/s. Find the range of the slug.

I solved it this way but I'm not sure if I did it correctly.

Vvi=(sin50)(3.37m/s)=2.581m/s
Vhi=(cos50)(3.37m/s)=2.166m/s

$$vertical:$$
$$v_i=2.581\frac{m}{s}$$
$$d=1.15m$$
$$a=g=9.81\frac{m}{s^2}$$
$$t=?$$

$$d=v_it+\frac{1}{2}at^2$$

$$1.15=2.581t+4.905t^2$$

$$4.905t^2+2.581t-1.15=0$$

$$t=\frac{-2.581+\sqrt{6.661561+22.563}}{9.81}$$

$$t=\frac{2.82497}{9.81}=.288s$$

$$horizontal:$$
$$t=.288s$$
$$v_i=2.166\frac{m}{s}$$
$$a=0$$
$$d=?$$

$$d=v_it$$
$$d=(2.166)(.288)=.6238m$$

so my answer is .6238m...but this doesn't seem correct at all. During the lab the slug went an average of 1.4 m each trial. Someone help me!

2. Dec 13, 2004

### derekmohammed

The reason you are getting wrong anwsers is because of the value you are using for the d. The formula you are thinking of is y(t)= y(intitial) + V(intial)t - 1/2 gt^2. By inputting 1.15 as d you are saying that the max height is going to be 1.15m. This is incorect. To find the max height H= (V(initial))^2(sinx)^2 / 2g (I derived this from the formula V^2 = V(inti)^2 + 2a(y-y(intial). Then plgu in the value from the height into the intial equation and solve for time. After soving for time use the standard d=vt.

OR!!!!
Use the formula
Range= (V(intial)^2)(sin2x) / g

Hopefully I have been helpful.

3. Dec 13, 2004

### Tide

You have the wrong equation for the height. The acceleration is downward so you want a = -g. With your equation the projectile will continue to rise forever - I don't think that's consistent with your observations. :-)