Does Eric's Leap Clear the Puddle?

[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.

 

[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.

 

[KingNothing]

Thanks Chen, I almost forgot t is just 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.

 

[ShawnD]

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

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.