finding maximum height of an object?

Question

I need help with this problem. A rock is thrown straight up with an initial velocity of 48 feet per second from an initial height of 64 feet above the ground. Its height h at time t is given be the formula h(t)=-16t^2+48t+64

1. find the time when the rock hits the ground.
I got this one 4 seconds, right?

2. find the maximum height of the rock?
this is the tough one? what should i do to solve it?

Answer 1

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

when time is -1 sec or 4 sec, the height is 0, and since time can't be negative, the answer is 4 sec.

2) to solve for this, you need to find the vertex: (-b/2a, f(-b/2a))
-b/2a = (-48 / -(16)2) = 1.5 sec
the hieght is equal to when you plug in the 1.5 sec for the time...

Answer 2

How To Find Maximum Height

Answer 3

RE:
finding maximum height of an object?
I need help with this problem. A rock is thrown straight up with an initial velocity of 48 feet per second from an initial height of 64 feet above the ground. Its height h at time t is given be the formula h(t)=-16t^2+48t+64

1. find the time when the rock hits the ground.
I got this one 4...

Answer 4

take positive Yaxis vertically upwards ano origin at the ground.:
1.y(0)_=64ft
u=48ft/s
a=-32ft/s^2
y=0
t+?
use y=y(0)+ut+1/2at^2
0=64+48t-16t^2
divide both sides by 16;
t^2-3t-4=0
t=4,-1
rejecting the negative answer, we get t=4s.
2.v=0,y=h
using v^2-u^2=2aS
0-48^2=-64h
h=48^2/64=48*3/4
=36 ft
this is the height abive the initial height from where the rock wasthrown up. So total maximum height
=36+64=100ft above the ground. ANS.

Answer 5

Max height is when upward velocity is zero, so we have
initial velocity minus deceleration to zero velocity or
48 - 32 t = 0 or t = 3/2. Substitute into first equation and we have

-16(3/2)^2 + 48 (3/2) + 64 = -16 (9/4) + 72 + 64 = -36 + 136 = 100

To check this result, add or subtract a very small amount from the time and recompute.

Answer 6

First the second part. The x-value of the turning point of all parabolas of the form y=ax^2+bx+c is x = -b/2a
Use that to find the time of maximum height.
t = -48/-32 = 1.5 sec.
Substitute to get the maximum height.
I will leave that to you.
Regarding your first question,
Solving 0 = -16t^2 + 48t + 64 (dividing by -16)
0 = t^2 - 3t - 4 = (t-4)(t+1)
The only applicable solution is t =4 as you have

Answer 7

For the best answers, search on this site https://shorturl.im/uVBHW

a) h '(t) = 80-32t = 0 for maximum t = 2.5 seconds h(2.5) = 100 ft b) 2.5+2.5 = 5 seconds

Answer 8

how you solve this depends on what class you're in... if this is a noncalculus-based physics class, then i guess you should just use the fact that (-b/(2a)) gives the t value of a the extrema of a quadratic function... using this you get

t = (-48)/(2*(-16))
t = (-48)/(-32)
t = 1.5 seconds

so, at time = 1.5 seconds, you should achieve your maximum height. this is just a matter of plugging 1.5 into the equation, so i leave it to you.

IF you've taken calculus, you should take the derivative of the f(t) function, then solve for zero, and find that t = 1.5. same answer, just more straightforward i think.

Answer 9

max height occurs when x = -b/2a = -48/-32 = 1.5
max heiht = -16(1.5)^2+48(1.5) + 64 = =100 ft.

Answer 10

h(t)=-16t^2+48t+64

So to find the time it hit's the ground we want h(t) = 0

or -16t^2 + 48t + 64 = 0

By the quadratic formula
(solution to ax^2 + bx + c = 0 is
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
)

we get
t = (-48 +/- sqrt(48^2 - 4(-16)(64))) / (2(-16))
= (-48 +/- sqrt( 2304 + 4096)) / -32
= (-48 +/- sqrt(6400)) / -32
= (-48 +/- 80) / -32
= 32 / -32 OR -128 / -32
= -1 OR 4

So the rock hit's the ground at 4 seconds.

Now; we also know that it would also have been at ground level at -1 seconds (ie had we thrown it up from the one ground, in 1 second it would have been at 64 feet), so, the total time from being at ground level to being back at ground level would have been 5 seconds (-1 to 4).

So the rock would have been at it's maximum height 2.5 seconds after leaving the ground.

We know that it would have left the ground at -1 seconds, so the maximum heiht would have happened at -1+2.5 or 1.5 seconds after it being thrown upwards from the 64 feet level.

So max height would be
h(1.5) = -16(1.5)^2 + 48(1.5) + 64 or
h(1.5) = -16(2.25) + 72 + 64
h(1.5) = -36 + 72 + 64
h(1.5) = 100