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A question about free fall, velocity, and the height of an object.

4

$begingroup$

A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.

a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?

b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?

I got the wrong answers when working on this.

To solve a):

$$s(t+2) - s(t) = 84$$
$$s(t) = v_0t+cfrac12at^2, v_0 = 5, a = 32$$
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
$$64t=10$$
$$t=cfrac58$$
$$5left(cfrac58right)+16left(cfrac58right)^2=9.375$$
$$h_0=109.375$$

To solve b):

$$100=-16t^2+7t+h_0$$
$$16=-16(t+2)^2+7(t+2)+h_0$$
now subtract the smaller constant from the larger
$$-84=-71t+7t-50$$
$$t=cfrac3471$$
$$100=-16left(cfrac3471right)^2+7left(cfrac3471right)+h_0$$
$$h_0=cfrac5056985041$$

However the answers are:
$a=cfrac647565$
$b=100$

What am I doing wrong?

calculus

share|cite|improve this question

asked 4 hours ago

JinzuJinzu

403513

$endgroup$

add a comment | 

4

$begingroup$

A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.

a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?

b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?

I got the wrong answers when working on this.

To solve a):

$$s(t+2) - s(t) = 84$$
$$s(t) = v_0t+cfrac12at^2, v_0 = 5, a = 32$$
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
$$64t=10$$
$$t=cfrac58$$
$$5left(cfrac58right)+16left(cfrac58right)^2=9.375$$
$$h_0=109.375$$

To solve b):

$$100=-16t^2+7t+h_0$$
$$16=-16(t+2)^2+7(t+2)+h_0$$
now subtract the smaller constant from the larger
$$-84=-71t+7t-50$$
$$t=cfrac3471$$
$$100=-16left(cfrac3471right)^2+7left(cfrac3471right)+h_0$$
$$h_0=cfrac5056985041$$

However the answers are:
$a=cfrac647565$
$b=100$

What am I doing wrong?

calculus

share|cite|improve this question

asked 4 hours ago

JinzuJinzu

403513

$endgroup$

add a comment | 

$begingroup$

A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.

a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?

b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?

I got the wrong answers when working on this.

To solve a):

$$s(t+2) - s(t) = 84$$
$$s(t) = v_0t+cfrac12at^2, v_0 = 5, a = 32$$
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
$$64t=10$$
$$t=cfrac58$$
$$5left(cfrac58right)+16left(cfrac58right)^2=9.375$$
$$h_0=109.375$$

To solve b):

$$100=-16t^2+7t+h_0$$
$$16=-16(t+2)^2+7(t+2)+h_0$$
now subtract the smaller constant from the larger
$$-84=-71t+7t-50$$
$$t=cfrac3471$$
$$100=-16left(cfrac3471right)^2+7left(cfrac3471right)+h_0$$
$$h_0=cfrac5056985041$$

However the answers are:
$a=cfrac647565$
$b=100$

What am I doing wrong?

calculus

share|cite|improve this question

asked 4 hours ago

JinzuJinzu

403513

$endgroup$

share|cite|improve this question

asked 4 hours ago

JinzuJinzu

403513

share|cite|improve this question

asked 4 hours ago

JinzuJinzu

403513

asked 4 hours ago

JinzuJinzu

403513

asked 4 hours ago

JinzuJinzu

403513

2 Answers
2

active

oldest

votes

1

$begingroup$

The error in a) is simple:

From $64t=10$ it follows $t=frac532 neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:

$h_0=100+5left(frac58right) + 16left(frac58right)^2=frac647564$

which is very similar to your answer key (I assume you mistyped the denominator).

In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.

share|cite|improve this answer

answered 3 hours ago

IngixIngix

5,097159

$endgroup$

add a comment | 

1

$begingroup$

the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$

share|cite|improve this answer

answered 3 hours ago

E.H.EE.H.E

16.1k11968

$endgroup$

add a comment | 

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1

$begingroup$

The error in a) is simple:

From $64t=10$ it follows $t=frac532 neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:

$h_0=100+5left(frac58right) + 16left(frac58right)^2=frac647564$

which is very similar to your answer key (I assume you mistyped the denominator).

In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.

share|cite|improve this answer

answered 3 hours ago

IngixIngix

5,097159

$endgroup$

add a comment | 

1

$begingroup$

The error in a) is simple:

From $64t=10$ it follows $t=frac532 neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:

$h_0=100+5left(frac58right) + 16left(frac58right)^2=frac647564$

which is very similar to your answer key (I assume you mistyped the denominator).

In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.

share|cite|improve this answer

answered 3 hours ago

IngixIngix

5,097159

$endgroup$

add a comment | 

$begingroup$

The error in a) is simple:

From $64t=10$ it follows $t=frac532 neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:

$h_0=100+5left(frac58right) + 16left(frac58right)^2=frac647564$

which is very similar to your answer key (I assume you mistyped the denominator).

In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.

share|cite|improve this answer

answered 3 hours ago

IngixIngix

5,097159

$endgroup$

share|cite|improve this answer

answered 3 hours ago

IngixIngix

5,097159

answered 3 hours ago

IngixIngix

5,097159

answered 3 hours ago

IngixIngix

5,097159

1

$begingroup$

the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$

share|cite|improve this answer

answered 3 hours ago

E.H.EE.H.E

16.1k11968

$endgroup$

add a comment | 

1

$begingroup$

the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$

share|cite|improve this answer

answered 3 hours ago

E.H.EE.H.E

16.1k11968

$endgroup$

add a comment | 

$begingroup$

the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$

share|cite|improve this answer

answered 3 hours ago

E.H.EE.H.E

16.1k11968

$endgroup$

share|cite|improve this answer

answered 3 hours ago

E.H.EE.H.E

16.1k11968

answered 3 hours ago

E.H.EE.H.E

16.1k11968

answered 3 hours ago

E.H.EE.H.E

16.1k11968