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# Physics speed/acceleration

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A rock is dropped from a bridge, When it's 1m from the water it takes 0.05 seconds to hit the water.

Calculate the height of the bridge.

(Air resistance is ignored)

May 28, 2018

#1
+28125
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First, use s = v1t + (1/2)gt2 to find v1, the velocity at s = 1m (g = 9.81 m/s2, t = 0.05 s).

Then use v12= 2gh to find the distance, h, from the top of the bridge to 1 m above the water (assuming rock is dropped with zero initial velocity).

Then height of bridge = h + 1 m

May 28, 2018
#2
+22884
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A rock is dropped from a bridge, When it's 1m from the water it takes 0.05 seconds to hit the water.

Calculate the height of the bridge.

$$\begin{array}{|lrcll|} \hline (1)& h &=& \frac{g}{2}t^2 \\ & t &=& \sqrt{\frac{2h}{g}} \\\\ (2)& h-1 &=& \frac{g}{2}(t-0.05)^2 \quad & | \quad t=\sqrt{\frac{2h}{g}}\\ & h-1 &=& \frac{g}{2} \left(\sqrt{\frac{2h}{g}}-0.05 \right)^2 \\ & h-1 &=& \frac{g}{2} \left(\frac{2h}{g} -2\cdot 0.05 \cdot \sqrt{\frac{2h}{g}} + 0.05^2 \right) \\ & h-1 &=& \frac{g}{2} \cdot \frac{2h}{g} -\frac{g}{2} \cdot 2\cdot 0.05 \cdot \sqrt{\frac{2h}{g}} + \frac{g}{2} \cdot 0.05^2 \\ & h-1 &=& h - 0.05\cdot g \cdot \sqrt{\frac{2h}{g}} + 0.05^2 \cdot \frac{g}{2} \\ & -1 &=& - 0.05\cdot g \cdot \sqrt{\frac{2h}{g}} + 0.05^2 \cdot \frac{g}{2} \\ & 0.05\cdot g \cdot \sqrt{\frac{2h}{g}} &=& 1+ 0.05^2 \cdot \frac{g}{2} \quad | \quad \text{square both sides} \\ & 0.05^2\cdot g^2 \cdot \frac{2h}{g} &=& \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2 \\ & 0.05^2\cdot g \cdot 2h &=& \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2 \\\\ & \mathbf{ h } & \mathbf{=} & \mathbf{ \dfrac{ \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2} {2\cdot 0.05^2\cdot g } } \\ \hline \end{array}$$

$$\text{Let g = \frac{9.81\ m}{s^2} }\\ \text{h the height of the bridge}:$$

$$\begin{array}{|rcll|} \hline h & = & \dfrac{ \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2} {2\cdot 0.05^2\cdot g } \\\\ & = & \dfrac{ \left(1+ 0.05^2 \cdot \frac{9.81}{2} \right)^2} {2\cdot 0.05^2\cdot 9.81 } \\\\ & = & \dfrac{ \left(1+ 0.05^2 \cdot \frac{9.81}{2} \right)^2} {2\cdot 0.05^2\cdot 9.81 } \\\\ & = & \dfrac{ 1.0122625^2} {0.04905 } \\\\ & = & \dfrac{ 1.02467536891} {0.04905 } \\\\ & = & 20.8904254619 \\ \hline \end{array}$$

The height of the bridge is  $$\approx 20.89\ m$$

May 28, 2018