Object Above Earth's Surface

magentarita · #1 $Old$ July 6th, 2009, 05:03 PM

An object is fired straight up from the earth's surface at 200 feet per second. Its height s off the ground at time
t is given by the function s(t)=-16t^2 +200t.

1) What is the initial height of the object?
2) When does the object hit the ground?
3) What is the domain of the object
4) What is the velocity, speed, and acceleration

artvandalay11 · #2 $Old$ July 6th, 2009, 05:20 PM

Quote:

Originally Posted by magentarita $View Post$

An object is fired straight up from the earth's surface at 200 feet per second. Its height s off the ground at time
t is given by the function s(t)=-16t^2 +200t.

1) What is the initial height of the object?
2) When does the object hit the ground?
3) What is the domain of the object
4) What is the velocity, speed, and acceleration

The initial height is the height when t=0, so evaluate s(0)

The object hits the ground when s(t)=0, or when -16t^2+200t=0

The domain of the object is t=0, until the object hits the ground, which you find in the previous part

The velocity is s'(t), the speed is the absolute value of the velocity, and the acceleration is s''(t)

yeongil · #3 $Old$ July 6th, 2009, 10:45 PM

Quote:

The object hits the ground when s(t)=0, or when -16t^2+200t=0

This probably does not need mentioning, but you'll get two answers for t, which makes sense, because at t = 0 the object is starting from the ground.

Quote:

The velocity is s'(t), the speed is the absolute value of the velocity, and the acceleration is s''(t)

Since the OP posted in the Pre-Calculus forum, we'll have to use something else.

The (standard?) position and velocity equations are the following:
$s(t) = s_0 + v_0t + \frac{1}{2}at^2$
$v(t) = v_0 + at$
where
$s_0$ is the initial height,
$v_0$ is the initial velocity,
$a$ is the acceleration,
$t$ is time,
$s(t)$ is the height at time t, and
$v(t)$ is the velocity at time t.

The OP asked in #4,

Quote:

4) What is the velocity, speed, and acceleration

but it is unclear which velocity and speed he/she is asking, because velocity/speed are always changing. Perhaps it's the velocity and speed when the object hits the ground again? The OP will need to clarify.

As for the acceleration, it's easy enough, just set the t squared coefficients equal to each other:
$\frac{1}{2}a = -16$
and solve for a.

01

magentarita · #4 $Old$ July 7th, 2009, 03:45 AM

Quote:

Originally Posted by artvandalay11 $View Post$

The initial height is the height when t=0, so evaluate s(0)

The object hits the ground when s(t)=0, or when -16t^2+200t=0

The domain of the object is t=0, until the object hits the ground, which you find in the previous part

The velocity is s'(t), the speed is the absolute value of the velocity, and the acceleration is s''(t)

This is helpful. Thanks.

magentarita · #5 $Old$ July 7th, 2009, 03:46 AM

Quote:

Originally Posted by yeongil $View Post$

This probably does not need mentioning, but you'll get two answers for t, which makes sense, because at t = 0 the object is starting from the ground.

Since the OP posted in the Pre-Calculus forum, we'll have to use something else.

The (standard?) position and velocity equations are the following:
$s(t) = s_0 + v_0t + \frac{1}{2}at^2$
$v(t) = v_0 + at$
where
$s_0$ is the initial height,
$v_0$ is the initial velocity,
$a$ is the acceleration,
$t$ is time,
$s(t)$ is the height at time t, and
$v(t)$ is the velocity at time t.

The OP asked in #4,

but it is unclear which velocity and speed he/she is asking, because velocity/speed are always changing. Perhaps it's the velocity and speed when the object hits the ground again? The OP will need to clarify.

As for the acceleration, it's easy enough, just set the t squared coefficients equal to each other:
$\frac{1}{2}a = -16$
and solve for a.

01

Thanks for the great details.