How would you do this velocity problem?

Question

If a ball is thrown into the air with an initial velocity of 104 ft/s, then its height (in feet) after t seconds is given by y= 104t^2-16t^2 . Find the velocity when t=4 .


Select the correct answer.

a) -9 ft/s
b) -29 ft/s
d) -24 ft/s
e) -19 ft/s

What would be the answer be, and how did you find it? Any help would be greatly appreciated!

Answer 1

There's something seriously wrong with your equation.

The proper equation, assuming that you're using earth's gravity, is:

y=104t-16.1t^2

differentiate w.r.t.

dy/dt = v = 104 -32.2t

now just plug in t=4 to get v(t=4). Notice that at t=0, velocity is that intial velocity. Note also the signs.

Answer 2

1) The position of the ball as a function of time is given as y(t) = k * t^2; k = 104 - 16 = 88

2) The velocity of the ball, @ t, equals dy/dt = 2kt

3) The velocity of the ball @ t=0, is zero; however, this contradicts the given 104 value -- The problem is improperly defined!!

4) Note that everything is nice if y(t) = 104t - 16t^2. In this case the correct answer is d)

Answer 3

Ur Eqn is WRONG

It must be y = 104t - 16 t^2

But it is not a necessary one.

Now apply this formula

v = u + gt

= 104 + (- 32)x 4 Where g = -32 , it is against the gravity


= 104 - 128

= -24 ft/s

Answer 4

Let height = h
h = 104t^2 - 16t^2
= 88t^2 feet
Let initial velocity = u
u = 104ft/s

t = 4,

So, h = 88 * (4)^2
= 88*16
= 1408 feet

Let final velocity = V
Let acceleration be a

h = ut + 1/2 at^2
1408 = 104*4 + (1/2 * a * 16)
1408 = 416 + 8a
416 + 8a = 1408
8a = 1408 - 416
a = 992/8
a = 124 ft/s^2

Now, V^2 = 2ah + u^2
V^2 = 2*124*1408 + (104)^2
= 349184 + 10816
= 360000
V = 600 ft/s

Answer 5

No sweat, just a typo,

y = 104t - 16t^2 (Gen form is y = y0 - 0.5gt^2)

Differentiate to obtain speed

v = v0 - gt

= 104 - 32*4

= -24 f/s

Answer 6

Let v = u - gt
Let s = ut - (1/2) g t²

s = 104t - 16t²--------(where s is distance)
ds/dt = 104 - 32t = v-----(where v is velocity)
v = 104 - 4 x 32------(after 4 seconds)
v = 104 - 128
v = - 24 ft / s-----------ANSWER d)

Answer 7

As x techniques 2 from the valuable or unfavorable area? It rather does make a distinction. If from the unfavorable, the numerator would be an relatively small neg. quantity and denominator an relatively small neg. quantity and because x^3 will reason the numerator to approach 0 quicker than (x-2) will, the shrink will approach 0. From the valuable, numerator will replace right into a vey small valuable and denominator an relatively small valuable yet, the effect would be an identical because of the fact the neg.. The shrink will back approach 0.

Answer 8

the equation given in the problem looks wrong.

Answer 9

That doesn't make any sense to me & I am pretty good at both physics and math...Did you forget a variable somewhere?