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A question about free fall, velocity, and the height of an object.
The Next CEO of Stack OverflowVelocity Question & AccelerationUp and Down Motion (Two objects meeting in time?)Velocity of a Ball When it Hits the GroundHeight and velocity of ball thrown verticallyRelated rates problem, rocket and observerThrowing a baseball on top of a cliffGiven initial conditions, find the maximum height reached by an object thrown upwards and its velocity on returning to the groundCalculus- Conceptual question about velocity.How does the sign of the acceleration depends on the direction of the distance choosen?Confusion on when velocity and acceleration are positive vs negative
$begingroup$
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
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
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
asked 4 hours ago
JinzuJinzu
403513
$endgroup$
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
calculus
asked 4 hours ago
JinzuJinzu
403513
asked 4 hours ago
JinzuJinzu
403513
asked 4 hours ago
JinzuJinzu
403513
asked 4 hours ago
JinzuJinzu
403513
asked 4 hours ago
JinzuJinzu
403513
403513
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 3 hours ago
IngixIngix
5,097159
$endgroup$
add a comment |
$begingroup$
the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$
answered 3 hours ago
E.H.EE.H.E
16.1k11968
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered 3 hours ago
IngixIngix
5,097159
$endgroup$
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.
answered 3 hours ago
IngixIngix
5,097159
answered 3 hours ago
IngixIngix
5,097159
answered 3 hours ago
IngixIngix
5,097159
answered 3 hours ago
IngixIngix
5,097159
5,097159
add a comment |
add a comment |
$begingroup$
the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$
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 $t=frac58$
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 $t=frac58$
answered 3 hours ago
E.H.EE.H.E
16.1k11968
$endgroup$
the solution of
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
should be $t=frac532$ not $t=frac58$
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
answered 3 hours ago
E.H.EE.H.E
16.1k11968
16.1k11968
add a comment |
add a comment |
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