could anyone possibly help me w/ y physics?

Question

a ball is thrown straight downward from a height of 75 meters. the initial velocity of the ball is 2 meters per second. (a) find the time it takes for the ball to reach the ground. (b) find the velocity of the ball just before it hits the ground.

Answer 1

lol i'm taking physics too i have a b in the class but i forgot how to do this XD there should be a formula for it in ur book

Answer 2

a lot of good people have already answered your question and they are correct

however, have really understood what they did?

you see darling, physics is much more enjoyable to learn if you could really get into the theories, and not just learning to use the formulas.

for example, your problem is a little tricky because you mentioned 'thrown'. in calculation of falling bodies, thrown is very much different from 'dropped'.

i will explained how.

a free falling body is the one we call 'dropped'. meaning there is no other force aside from gravity that makes the object go down to the earth.

'thrown' on the other hand indicates that aside from gravity, another force is given to the object for it to fall to the ground as if throwing a baseball.

you can see this difference inthe formulas we use.

in free falling body:
height is equivalent to one half of the gravity pull (8.9m/sec^2) multiplied to the time (exponentially twice) it took to hit the ground

simply written as: h = 1/2 g t^2

and velocity is formula is written as: v^2 = 2 g h

these simple formulas are enough to calculate any given problems in free falling bodies.

but when it come to forced acceleration, the additional 'initial velocity' must be taken into account.

h = Vo + 1/2 g t^2

v^2 = Vo^2 + 2 g h

we simply added Vo to the formulas - this is original or initial velocity.

h is most of the time written as "s" in the formulas but they mean the same thing - 'height'. do not be confused.

if you think about it or imagine a little, using this formulas gives you a glimpse of how an object is reall travelling in air.

isn't it great ?!

enjoy your studies !

Answer 3

D = 75
D = 1/2 a times T(squared) + initial velocity T
75 = 1/2 9.8 m/s/s T(squared) + 2 m/s T
0 = 4.9 T(squared) + 2 T - 75
solve for T (in seconds) with quadratic A=4.9, B = 2, C = -75
Ignore the negative answer.
V terminal = 2 + 9.8 T (in m/s)

Answer 4

y = vit + .5gt^2
75 = 2t + 4.9 t^2
0 = 4.9t^2 + 2t - 75, a simple quadratic.

No calculator handy, but that should help a little for part a.

vf = vi + at
vf = 2 + 9.8t, using t value from a.

Answer 5

From the equations of linear motion,we have
(a)s=ut + 1/2at^2
Given that s=75m,u=2m/s,a=10m/s^2
75=2t+1/2(10)(t^2)
75=2t+5t^2
So,5t^2+2t-75=0
Factorise it,u wil get t=3.68s or t=-4.08s
But t can't be negative,so t=3.68s

(b)Another equation

v=u+at
=2+10(3.68)
=38.8m/s

Answer 6

v^2 = U^2 +2as u = 2 m/s a = 9.8 m/s^2, t = (V-U)/a

V^2 = 4 +2x9.8x75 = 1474, V = sqrt(1474) = 38.39 m/s

t = (38.39-2)/9.8 = 3.71 s