Don't really understand free fall?

Question

question:

A man leans over the edge of a cliff and throws a rock upward at 30 m/s. Neglecting air resistance, what is the rock's speed 9 seconds later?

so from what i understand, the rock is going upwards at a speed of 30m/s, decreasing by 10 (due to gravity) every second, until it reaches 0, the top of it's path. So at 3 seconds, the speed is 0 correct? then the speed goes from 0 to -10 by second 4, -20 by 5s, -30, by 6s, and -60 by 9seconds? would the answer to that problem be -60 then??

is there some sort of formula to this so i don't have to draw it out each time?

also, how would i figure out the answer to these two problems, they are similar, but slightly diffrent, and the slight diffrence gets me lost:

If a projectile is fired straight up at a speed of 36 m/s, how much time does it take to reach the top of its path? Hint: Final speed is zero.

A ball is thrown upwards. Neglecting air resistance, what initial upward speed does the ball need so that it reaches the top of its path in 9 seconds? Hint: Final speed is zero.

i don't really want the answers as much as i want an explanation or possible formula so i can figure out how to do this on my own. thanks!

Answer 1

Your approach is correct.  The value of g is not 10 m / sec^2 but it is 9.81 m / sec^2.

The formulae to be used are

v = u + at

s = ut + 1/2 a t^2

and

v^2 = u^2 + 2as

where u is the initial velocity

a is the acceleration

t is the time taken

and v is the final velocity

When an obejct is thrown up, it decelrates at eht arate of 9.81 m / sec^2 and so the value of a for such a situatin should be taken as

-9.81 m / sec^2

For the projectile question,

v = u + at

here the finlal velocity, v = 0

initial velcoity u = 36 m / sec

a = -9.81 m / sec^2

0 = 36 - 9.81t

or t = 36 / 9.81 seconds

For the second question,

v = 0

a = -9.81 m / sec^2

t = 9 secs.

u  = ?

0 = u - at

0 = u - 9.81 x 9

u = 9.81 x 9 = 89.01 m / sec

The velocity with which a ball has to be throw upwards so that it takes 9 secs to reach the top position is 89.01 m / sec.

Hope the above is clear

Answer 2

In this one, you are relating three variables, speed, acceleration and time. These are related by the linear formula of V(t) = at + Vi.  a is deceleration due to gravity, t is time, Vi is initial velocity.  It is a linear relationship, if that term means something to you (if graphed would be a line), so as time goes on, the acceleration has more and more effect on the speed, just as you said.

For the one involving solving for time to zero speed, you use the same equation, and set at + Vi equal to zero, then solve for t.

For the one involving 9 seconds of flight, slowing to 0, you set the correct a and t (9 sec.) in formula, then set  at + Vi equal to 0 and solve for the Vi.

That is a key in physics, to understand the relationship between the variables.

Answer 3

let me simplify these essays for you, because you obviouslyt get the concept :)

your equation is V=V0 + A(T)

velocity = initial velocity + acceleration x time

velocity = 30 m/s + (-10m/s x 9)

              = 30 m/s - 90 m/s

              = -60 m/s

so you are correct.

Answer 4

Your reasoning showed that you understood the concepts very well. For more practical purposes you can use formulas.

In relation to the velocity of a body moving under the effect of a constant force, you can apply the formula

v = at + v0 (v0 is initial velocity), or

v - v0 = at, or

t = (v - v0)/a, or

a = (v - v0)/t

The formula v = at + v0 is obtained by derivating the position formula

x = at^2 + v0t + x0