velocity? is that like sincity?

Question

Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
•16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
•v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
•s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
a)What is the function that describes this problem?

b)The ball will be how high above the ground after 1 second?



c) How long will it take to hit the ground?





d) What is the maximum height of the ball? What time will the maximum height be attained?

Answer 1

a) they give you everything you need. v0 = 32, s0 = 0, so

s(t) = -16t² + 32t

b)

s(1) = -16 + 32 = 16

c) The ball hits the ground when s(t) = 0 (the second time).

0 = -16t² + 32t = -16t(t-2) so t=0 or t-2 = 0, t=2.

So the ball hits the ground when t=2.

d) The maximum height is attained at the vertex of the parabola.

y - 16 = -16(t² - 2t + 1)
y - 16 = -16(t - 1)²

So the vertex is (1, 16), that is, at time t = 1, when the height of the ball is 16.

Answer 2

a. If v0 = 32, then the equation is:

s = 32 t -16 t^2.

b. At t = 1, s = 32 (1) - 16 (1)^2 = 32 - 16 = 16 feet.

c. The ball will hit the ground when s = 0:

0 = 32t - 16t^2

0 = 16t(2 - t)

16 t = 0 or 2 - t = 0

t = 0 or t = 2.

This means that the ball will be at the ground when it is thrown (at t = 0) or 2 seconds after it is thrown (when it finally lands).

d. The equation is that of a parabola. It will be at its highest point halfway between when it is thrown and when it lands. This happens at t = 1 second.

As indicated in part (b), it will be 16 feet above the ground.

Answer 3

b. S=-16*1^2 + 32 * 1 + 0
S= 16 ft
d max if ds/dt=0 ..... -32t+32 = 0 ... t=1
S = 16 ft
c. T= time to hit the ground = 2 times the time ball reach max height that is 2 * 1 = 2 sec

a. ??? function height versus time ???

Answer 4

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Answer 5

i usually use meter.
32ft/s--->9.7536m/s

b) x=1/2at^2+Vit+xi
x=(1/2)(-9.8m/s^2)(1s)^2+(9.7536m/s)(1s)
x= -4.9m+9.7536m
x=4.8536m (or 15.92ft)

c) Vf=at+Vi
-9.7536m/s=(-9.8m/s^2)t+9.7536m/s
-19.5072m/s=(-9.8m/s^2)t
t= 2 s

d) Vf^2=2ad+Vi^2
when it reaches the highest, its final velocity will be 0m/s
(0m/s)^2=2(-9.8m/s^2)d+(9.7536m/s)^2
-95.1327m^2/s^2= (-19.6m/s^2)d
d=4.8537m (or 15.92ft)

the time it reaches the highest point is half the time it reach the ground so t= 1s
or 0m/s=(-9.8m/s^2)t+9.7536
-9.7536m/s=(-9.8m/s^2)t
t=1s