A question about free fall, velocity, and the height of an object.
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}$$b=100$
What am I doing wrong?
$\endgroup$ 02 Answers
$\begingroup$The error in a) is simple:
From $64t=10$ it follows $t=\frac5{32} \neq \frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:
$h_0=100+5\left(\frac58\right) + 16\left(\frac58\right)^2=\frac{6475}{64}$
which is very similar to your answer key (I assume you mistyped the denominator).
In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.
$\endgroup$ $\begingroup$the solution of $$\left[5(t+2)+16(t+2)^2\right]-(5t+16t^2)=84$$should be $t=\frac{5}{32}$ not $t=\frac{5}{8}$
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