Question

maximum and minimum problem

Answer 1

by differentation?

Answer 2

idk. maybe? =) this is an advanced topic in quadratic equation

Answer 3

oh mechanics, so u=70, a or g = -9.8, at max point its essentially not going up or down, so v=0. then you need s, so use v^2 = u^2 - 2as

Answer 4

sorry that should be + 2as

Answer 5

this equation is not applicable because its a projectile motion..

Answer 6

Ok well 160 ft/s is approximately 48.5m/s

Answer 7

v=u+at gives your answer of 5s

Answer 8

This is really a physics problem often seen in calculus courses, but it can be reduced to a problem involving quadratic equations.

DemandAndSupply is on the right track. Still, much depends on whether or not you've heard and/or read about problems involving gravity.

If " a " represents acceleration and is constant, then velocity is simply \[v = v _{0}+at\]
where t represents time.

In a gravity problem (in the English system of measurement), a = g = -32.2 feet/(sec^2).

If "an object is shot vertically upward with an initial velocity 70 feet per second, find when the object reaches its maximum elevation," your initial velocity is 70 feet/sec:\[v _{0}=70 ft/\sec\], and your acceleration is

Answer 9

\[a = g =\frac{ -32 feet }{ \sec ^{2} }\]

Answer 10

and so you can calculate the time required for your object to reach its maximum height by substituting these values into\[v = v _{0} - g*t\]

Answer 11

since the object actually stops moving momentarily when it reaches its max height, let v = 0.

Solve\[0 = -32.2\frac{ ft }{ \sec ^{2} }+70\frac{ ft }{ \sec }\]

Answer 12

for t.

Answer 13

Sorry, that should be

"solve 0=-32.2 t + 70 for t." (I accidentally omitted the t.)

Answer 14

Thank you so much!!

Answer 15

My pleasure!

You'll still need to calculate the max height reached by this object.

Solve the previous equation for t. Then substitute this value of t into the position equation

\[s=s _{0}+\frac{ 1 }{ 2 }g*t ^{2}+v _{0}t.\]

Answer 16

Please note: the formula shared with you by @DemandAndSupply will also provide you with the max ht. My method is longer (a lot longer), but I wanted to explain where the formulas come from.

DemandAndSupply's formula,
\[v ^{2}=v _{0^{2}}+2as\]

Answer 17

can be solved for s:

Answer 18

\[s=\frac{ v ^{2}-v _{0}^{2} }{ 2a }\]

Answer 19

Then calculate the max height by substituting the following values:

v=0
\[v _{0}=70 ft/\sec\]

a=g=-32.2 ft/(sec^2)

Answer 20

@mathmale This guy has excellent rigorous knowledge. The way he is showing you is a more effective way to learn the subject as you will find it easier to apply these methods to future problems.

Answer 21

@DemandAndSupply: Thank you very much. I'd not thought of the formula you'd provided (although I actually taught physics in 1969!).

Answer 22

Well that can be forgiven as that was a while ago! Fortunately i only studied applied mechanics last year