this is another math question that i couldn't possibly solve after trying to figure it out so please help!!

Question

this is the math question and it has to do with quadratic equations(i think) given to me as part of my 10th grade math homework. It's not long at all, but very hard!

A ball is thrown directly upward from ground level with an initial speed of 80ft/s. How high will it go? When will it return to the ground??
Thanxx for helping me in advance!

Answer 1

Don't believe Dalton. You don't need the mass, and no one ever deals with air resistance, so you don't need bumps.

Gravity acts on all masses equally (the force is greater for larger masses, but that is exactly cancelled out by their momentum). You do need gravity, which as I recall is 32f/s^2.

Initial speed upward is 80ft/s. So the speed after t seconds is 80 - 32t. Solve for 0 to get the time to the highest point: 80 - 32t = 0, t = 80/32 = 5/2. It takes this same amount of time to come back down so the total time in the air is 5 seconds.

As for height I am not sure what you know and are supposed to use. The speed is 80 - 32t, so the height is -16t^2+ 80t + C, 0 at t = 0 so C = 0. At time 5/2 this is -16*(5/2)^2 + 80*5/2 = 100 feet.

Oh, Longman's explanation of the maximum height is very good. The speed is linear so the average speed is (fastest + slowest)/2, figuring the maximum height this is (80 + 0)/2 = 40, for 5/2 seconds = 100.

Answer 2

Three people attempted to answer this question. They failed. You don't need the mass. Gravity does not depend upon the mass. You always ignore the wind in this type of thing. There are two parts of this Question. The ball is thrown upwords with 80ft/sec. Gravity acts as a negative accelration and the ball will keep going up until the gravity reduces the speed to zero. You need to use two equations. First you have to find the time (t) till the final velocity is reduced to zero. v=u+ft. v=0 and u=80. f=-g. You should know g or you can't figure out the answer. If you can't solve this to find t you should not be in the 10th grade. Use this t to find the height it will go to. h=ut+ft*2 where u=80 and f=-g. Again if you can't solve this simple quation you should not be 10th grade.

The ball will return to the ground exactly the same time it took to go up with the final spped the same speed it had when it was thrown because it has the same accelration. g, the same height and the same time.

Sorry I can't help you cheat the system by doing the calculations.

Answer 3

80 ftper second =initial velocity
-32ft/sec = sped of gravity
we know that a basic foruma for this is 1/2gx^2+initialvelocity*x +initial height
thus position x would be

1/2(-32)x^2+80x+0=p(x)
-16x^2+80x+0=p(x)
A) how high will it go?
We know this function has a maxinum point, since it is an upside down parabola.
to find its vertex, use -b/2a
-80/-32= 2.5
when x is 2.5, the value of the position wuold be -100+200=100 ft
To check my work (don't worry abuot this part)
-32x+80=0
x=2.5 Yes
B) It will return to the ground when p(X)=0
-16x^2+80x=0
-16x(x-5)=0
when x = 5, it will hit the ground. (5 seconds)

Answer 4

Sorry, I can only work with SI units. Thus instead of giving numeric answer, please just accept my algebraic expressions.

Suppose your initial velocity is u. The gravity is g, which decelerates the ball's speed to zero at its highest point. Thus, the time takes to the highest point is u/g. The ball will take the same length of time to get back to the ground, i.e. 2(u/g).

The average speed of the ball from the ground to the highest point is u/2. Thus the distance from the ground to the highest point is u/2 * u/g = u^2/2g.

Answer 5

I always had to be what seemed like double the effort of my other classes into math. Just keep studying it; don't give up. Study twice as hard. But make sure you get good sleep every night because math makes no sense when you're tired you can make stupid mistakes.

Answer 6

Here's the answer to the first part. Using v^2-u^2 = 2fs, where v=final velocity, u=initial velocity, f=acceleration (gravity) probably 32ft/sec and s= distance, you get the answer 100ft.

Answer 7

Let V0 = intial speed = 80ft/s
gravity = g = 32.15 ft/s^2
mass = m
mechanical energy = Em
max height = h

Em ground = Em apex
m*g*h +0.5*m*v^2 = m*g*h + 0.5*m*v^2
0.5*m*v^2 = m*g*h
0.5*v^2 = g*h
height = v^2 / (2*g)
= 80^2 / (2*32.15)
= 99.5 feet

Vt=V0-g*t
0 =80 - 32.15 * t-to-max
t-to-max = 80/32.15
= 2.5 sec
t-to get back to ground = 2 * time to max
= 2*2.5 sec
= 5 sec

Answer 8

v^2-u^2=2gs
v=0,u=80ft/sec,g=32ft/sec^2
s=-80*80/2*32=100ft
It will go up 100ft
s=ut+1/2*g*t^2
s=100,u=o,g=32ft/sec^2 t=?
t^2=2*s/g=2*100/32=100/16
t=10/4=2.5sec
It will come down in 2.5sec

Answer 9

There should be a formula for this. Would you post it please? You have to find the maximum point of a parabola which will be your height. Once you find the height you can find the time, but I don't remember the formula.

Answer 10

i need to know the mass of the ball, wether there is wind, and the force of gravity.
if the ball has bumps(like a basket ball) its even harder

if you dont have this info, the question is unanswerable.