math question...?

Question

if a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 48ft/sec, its height after t seconds is s(t) = 64 + 48t - 16t^2. What is the maximum height the ball reaches? ___
What is the velocity of the ball when it hits the ground (height 0)

Answer 1

Given s(t) = 64 + 48t - 16t^2
s '(t) = -32 t + 48

s '(0) = 48 (indeed the initial velocity is 48 ft/s!)
s '(t) = 0 (this is the instance when the ball reaches its max height when its velocity os zero)
This means t = 48/ 32 =1.5.

Therefore s(1.5) = 100 ( maximum height from the ground)

If you set s(t) = 0 ( when the height is zero)
then you'll get t = 4 sec.

The velocity when the ball reaches the ground is s '(4) = -80 ( 80 ft/sec going down)

Answer 2

We know that Velocity V = dS/dt
i.e. V = d/dt(64+48t-16t^2
or V = 48-32t
But at maximum height V = 0
i.e. 48-32t = 0
or t = 48/32 = 3/2 seconds
Putting the value of t in the given equation
S(3/2) = 64+48*(3/2)-16*(3/2)^2
i.e. S(3/2) = 64+72-36 = 100 ft
Hence the maximum height reached by the ball is 100 ft.
It is clear that when the ball reaches a height of 64 ft while falling, its velocity is 48 ft/sec. Now, taking this as the initial velocity (u), the velocity of the ball when it hits the ground (v), can be found using the relation
v^2-u^2=2gS, where g is the acceleration due to gravity and S = 64 ft.

Answer 3

the expression above gives the height as a function of time, i.e. h(t)

so to find the maximum, first notice that the "a" term is negative...
then that means that the vertex of the parabola is the maximum, so to find the vertex, find [-b/2a, f(-b/2a)]

b/2a = -48 / -16(2) = 48/32 = 3/2

f(3/2) = 100

So the maximum height occurs at 1.5 sec. The max height is 100ft...


When height =0, the time is...

0 = 64 +48t - 16t^2
-t^2 +3t +4 =0
-(t-4)(t+1) = 0

Since time cannot be negative, the time is 3 sec for the entire trip, for the part just after the max height, subtract the 1.5 off...
So T = 2.5 sec
Use 100 ft as x, because we want the Vo to be 0 (the velocity at the maximum height is 0ft/sec)

t= 2.5 sec
Vo=0 ft/s
V=??
x= 100 ft

X/t = (V+Vo)/2
200/2.5 = V
V = 80 ft/sec

Answer 4

As per Newtons Law Eqn S=ut-1/2ft^2 , as you show in your Euation u= 48,
There for using V^2= U^2 - 2fs
V=0 at max height
0= 48x48 - 2x32xh ( putting f = 32 & S= h )
Hence , h = 48x48/64 = 36 feet
So the ball reaches the height of 64+36 =100 feet.
2nd part U=0 h= 100 feet
hene using v^2 = u^+ 2fh
V^2 = 0 + 2x32x100 = 6400 ( U= 0 initial Velocity.
Hence V = 80 feet /second

Answer 5

So to solve the maximum height, start plugging in times

1 sec = 96ft
2 sec = 96ft
3 sec = 64ft

So somewhere between 1 and 2 seconds it reaches its max height

1.5 seconds = 100ft

So max height = 100 ft

Now using that answer we can solve for the velocity of the ball when it hits the ground. But we need one more piece. At what point does it hit the ground.

So must solve for s=0

Factoring the equation we get (8-2t)(8+8t)

One of those parts must = 0

(8+8t)=0 t=-1, not a real point of time

(8-2t) = 0, t=4, so at 4 seconds the ball hits the ground.

So it took 1.5 seconds to reach its highest point 100ft. It then took 2.5 more seconds to hit the ground.

So velocity = 100ft/2.5secs = 40ft/sec

Answer 6

To maximize the height, differentiate s(t) with respect to t (which gives you the velocity v(t)), set the derivative equal to zero, and solve for t. Then put this t-value into the equation for s(t). The result is the maximum height.

To answer the second question: set s(t) = 0 and solve for t (you will get two solutions, one of which is negative; obviously you want the positive solution). Then plug that t-value into the velocity equation you will have obtained for the first question; that will give the velocity when the ball hits the ground. Your answer will be negative; this means that the velocity is downward.