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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):

$$s(t+2) - s(t) = 84$$ $$s(t) = v_0t+\cfrac{1}{2}at^2, v_0 = 5, a = 32$$ $$\left[5(t+2)+16(t+2)^2\right]-(5t+16t^2)=84$$ $$64t=10$$ $$t=\cfrac{5}{8}$$ $$5\left(\cfrac{5}{8}\right)+16\left(\cfrac{5}{8}\right)^2=9.375$$ $$h_0=109.375$$

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 $$-84=-71t+7t-50$$ $$t=\cfrac{34}{71}$$ $$100=-16\left(\cfrac{34}{71}\right)^2+7\left(\cfrac{34}{71}\right)+h_0$$ $$h_0=\cfrac{505698}{5041}$$

However the answers are: $a=\cfrac{6475}{65}$ $b=100$

What am I doing wrong?

Jinzu
  • 819

2 Answers2

1

the solution of $$\left[5(t+2)+16(t+2)^2\right]-(5t+16t^2)=84$$ should be $t=\frac{5}{32}$ not $t=\frac{5}{8}$

E.H.E
  • 23,280
1

The error in a) is simple:

From $64t=10$ it follows $t=\frac5{32} \neq \frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:

$h_0=100+5\left(\frac58\right) + 16\left(\frac58\right)^2=\frac{6475}{64}$

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.

Ingix
  • 14,494