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# From what height must a ball be tossed for it to strike the ground with a velocity of -136 ft/sec?

from what height must a ball be tossed for it to strike the ground with a velocity of -136 ft/sec?

A ball is thrown upward from the surface of the earth with an initial velocity of 96 ft/sec. What is the maximum height it reaches?

### 3 Answers

- PaulLv 78 years agoFavorite Answer
Assume acceleration due to gravity is -32 ft/sec^2

v^2 = u^2 + 2as but for simplicity I am going to assume u^2 = 0

v^2=2as

(-136 ft/sec)^2 = 2 (-32 ft/sec^2)(s)

18496 ft^2/sec^2 = 2 (-32 ft/sec^2)(s)

= -289 ft.

So the ball has travelled 289 feet down from the point at which it was tossed.

I normally make a point not to answer second questions as I believe people should get one question for their five points and asking multiple questions for five points is cheating the answerers out of their two points and one person out of their 10 but I'm in a very good mood so I'm going to give you the other one.

Again v^2 = u^2 + 2as but this time v^2 = 0 (which happens at maximum height, the ball has to stop going up before it can go down and in that interval its velocity is zero). We know the initial velocity is 96 ft/sec. Again for ease of maths we're going to assume acceleration under gravity is -32 feet per second squared.

0 = (96 ft/sec)^2 + 2 (-32 ft/sec^2) (s)

0 = 9216 ft^2/sec^2 -64 ft/sec^2 (s)

-9216 ft^2/sec^2 = -64 ft/sec^2 (s)

-9216 ft^2/sec^2/-64 ft/sec^2=s

144 ft = s

Hope this helps.

Please be nicer to the community by asking one question at a time.

- 8 years ago
nw, when the ball is comin down from max heigth

u= 0ft/s, v= 136ft/s, g ≈ 32ft/s

so,

v^2 = u^2 +2as

(136)^2 = 2(32) h

h=289 ft

nw, to distance covered in upwrd direction whn it is tossed,

here, u = 96ft/s , v=0ft/s

0^2 = (96)^2 - 2(32)h

h = 144ft

so, the ball should be tossed at a height 289-144= 145ft abv the ground so that it attains the speed of 136ft/s while hittin the ground

also the max height attained by the ball would b 289 ft

- wolskyLv 44 years ago
Use the equations of action. shall we assume that down is the powerful y course. g ~ 32 ft/s^2 v(t) = v0 + g t dy = a million/2 g t^2 we've 136 = 0 + g t --> t = 136 / g = 4.25 s dy = a million/2 g t^2 = a million/2 g (4.25)^2 = 289 ft