If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 32 ft/sec?

Question

What is the maximum height the ball reaches?

Answer 1

Alec113 is right.

If you were on the ground, and threw a ball up at 32 ft per sec, it would immediately begin decelerating at 32 ft/sec every second.

So after one second it would have decelerated by 32 ft per sec, and STOPPED. :-)

D = 1/2 * A * t^2,
so ...
1/2 * 32 * 1*1

16 feet.

that's 16 feet higher than the building!

so 64 feet high.

in your text you will have a formula using initial velocity and acceleration that would also tell you the answer.

also see the kinematics section of the citation below (try the first formula describing constant acceleration, and you will get the same answer for x final.)

x initial is 48 feet, (tho you could change the x's to y's to make it clearer that it is vertical feet we are discussing),

v initial is 32 ft/sec,

v final = 0 (when it stops and starts back down)

Xf = Xi + (Vi + Vf)/2
Xf = 48 + (32 + 0)/2
Xf = 48 + 16
Xf = 64

Good luck!

Answer 2

The maximum height is reached when the ball`s velocity is 0 (and then starts falling).
So imagine at first that you`re at ground level, and get that height (with the initial velocity given, and gravity acceleration slowing the ball).
Then you have to consider that the ball was actually thrown from the roof of a building. That`s done adding 48 foot to the maximum height you got before.

In other words, you have to set a new coordinate system in the roof of the building and get the maximum height. After that, you have to express what you`ve got, in terms of the original coordinate system at ground (adding the building`s height)

Answer 3

s = (v^2 - u^2)/2g.

Here, s = displacement, the concerned value
v = final velocity = 0
u = initial velocity = 32
g = acc. due to gravity = -32

It gives s = 16 ft.

Therefore the maximum height the ball reaches as observed from the ground is 48+16 = 64 ft.

Answer 4

i assume that the equation is: s(t) = 40 8 + 64t - 16t² enable s(t) = s'(t) = v(t) Take a spinoff of s(t), which provides: v(t) = sixty 4 - 32t enable v(t) = 0. Then, resolve for t to be sure the time the place the ball reaches the optimal top. 0 = sixty 4 - 32t this grants: t = 2 replace that to s(t) with t = 2 to be sure s(2). s(2) = 40 8 + sixty 4(2) - sixteen(2)² s(2) = 40 8 + 128 - sixty 4 for this reason, you have: s(2) = 112 feet ? optimal top at t = 2. 2nd section: to be sure the cost of the ball whilst it hits the floor, enable s(t) = 0 and resolve for t. Then, replace that to v(t) with t values. right here, we've: 0 = 40 8 + 64t - 16t² 0 = sixteen(3 + 4t - t²) stick to quadratic formulation... t = (-4 ± ?(4² - 4(-a million)(3)))/2(-a million) t = (-4 ± ?(sixteen + 12))/-2 t = (-4 ± ?28)/-2 this grants: t ? 4.sixty 5 seconds Reject t ? -0.sixty 5 seconds on account which you may no longer have unfavourable length. finally, replace t with (approx.) 4.sixty 5 seconds to v(t). v(4.sixty 5) = sixty 4 - 32(4.sixty 5) for this reason, we've: v(4.sixty 5) ? -eighty 4.sixty six feet/sec i wish this helps!

Answer 5

I dont think you have enough information to answer this question you either need

a) a formula to plug those numbers into
b) the force the ball was thrown
c)the number of seconds the ball was traveling upward until it began its descent

Answer 6

omg. physics. i love it.
well....
x(initial)=48, v= 32, g= -9.8, x=?

and use one of those formulas. but i cant remember it. and then solve for x... im guessing.
i could be wrong. HAHAHAHA.