1. Using the y -component of the initial velocity from Q#2, find the time to ris

Question

1.

Using the y-component of the initial velocity from Q#2, find the time to rise to the highest point. Show us in the figure (either one) where that highest point (apex) is (it is not H!) Hint: Begin your work with the following equation:

vy =v0y +ayt1 (Whatisay here?Whatisvy attheapex?)
Your final answer for t1 should only involve the following variables: v0, g, and ?.

2.

You can now find the height of the fish above the ground at its highest point, the apex. Show that this maximum height (again, this is not H!) is given by the following equation:

h = (v0 sin?)2
max ?? g , where the ?? means, perhaps, there is some coefficient of ?g? down there

(Is it 1? A number? Maybe a sign? Work it out for us, get the formula fixed up!)

3. Show us that the horizontal distance ?d? the fish jumped in the figure is simply d = v0 cos? (t1 + t2 ) Explain your reasoning

4.At last, some numbers! Suppose ? =60 deg, and lets suppose our fish can leap at an initial speed of 1.5 m/s For simplicity, let?s assume the second bowl is not any higher than the first, so set H=0. How far away, horizontally, can our little fish jump? (In other words, solve for d) Do these numbers seem physically reasonable for a goldfish?

A goldfish, in a desperate play for freedom, launches itself with initial velocity v0 at angle of with respect to horizontal. In the figure, the dashed line shows the path of the "fish projectile. The fish lands in a second bowl, splashing down a height H above where it started. It lands a horizontal distance "d" from where it launched C13 Image from http://enyonam.com/?attachment id-34 Here is the same physics, in a simplified diagram. The initial velocity vector is now shown 0 +X

Explanation / Answer

1. vy = v0y + ay t1

at max height vy = 0

0 = v0 sin theta - g t

t = v0 sin theta/g

2. v^2 = v0^2 + 2 a y

0 = (v0 sin theta)^2 - 2 g y

y = v0(sin theta)^2/(2g)

3.velocity in the x direction is constant

so ax = 0

so x = v0 cos(theta) t = v0 cos(theta) (t1 + t2)

4. Range = v0^2 sin(2 theta)/g = 1.5^2*sin(120 degrees)/9.81= 0.199 m

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