# Check my numbers?

1. Mar 24, 2004

### KingNothing

Sorry, this is extremely basic for most of you so thanks for even reading this...

"Eric is running to school and leaping over puddles as he goes. From the edge of a puddle 1.5 m long, he jumps .2 m high off the ground with a horizontal velocity of 3 m/s. WIll he land in the puddle?"

I said: .2=4.9t^2
.04...=t^2, t=0.2 s
0.2*3=0.6, 0.6 < 1.5 therefore, yes he will hit the puddle.

2. Mar 24, 2004

### Chen

The total time of the jump is actually $$2t$$, so the horizontal distance is 1.2 meters. Still, he won't make it over the paddle.

3. Mar 24, 2004

### KingNothing

Thanks Chen, I almost forgot t is jsut for one direction.

I have a question on this one too:

"A golf ball is hit, it travels 155 yards before hitting the ground 3.2 seconds later. What was it's initial velocity?"

Now, I found the height to be about 12.5 yards, but I don't know where to go from there.

4. Mar 24, 2004

### ShawnD

Just make 2 simultaneous equations.
I'll write the distances x3 (3 feet in a yard i think).

horizontal distance (solve for velocity):

$$d_x = V_xt$$

$$d_x = Vcos(\theta)t$$

$$V = \frac{d_x}{tcos(\theta)}$$

vertical distance (solve for velocity):

$$d_y = V_yt + \frac{1}{2}at^2$$

$$0 = Vsin(\theta)(t) + \frac{1}{2}at^2$$

$$V = \frac{-at}{2sin(\theta)}$$

Now if I did those right, this should work. Make the equations equal to each other and try to solve for theta (the angle)

$$V = V$$

$$\frac{d_x}{tcos(\theta)} = \frac{-at}{2sin(\theta)}$$

$$tan(\theta) = \frac{-at^2}{2d_x}$$

$$tan(\theta) = \frac{-(-32.2)(3.2)^2}{2(465)}$$

$$\theta = 19.52$$

Now back to the first equation

$$V = \frac{d_x}{tcos(\theta)}$$

$$V = \frac{465}{(3.2)cos(19.52)}$$

$$V = 154.17 \frac{ft}{s}$$

So the answer would be 154.17 ft/s at 19.52 degrees from the horizontal.
Remember how to solve for simultaneous equations, it will come back to haunt you.