This article aims to find the **time it takes the workers to reach the piano before it reaches the bottom**Â of the ramp. This **article uses the concept**Â of determining the **acceleration due to gravity**Â and the **length of the ramp. Gravitational acceleration**Â is the**Â acceleration**Â gained by an object due to the **force of gravity**. Its SI unit is $ \dfrac{m}{s ^ { 2 }} $. It has both magnitude and the direction, so it is a **vector quantity**. **Gravitational acceleration**Â is represented by $ g $. The **standard value**Â of $g$ on the earth’s surface at **sea level**Â is $ 9.8\dfrac {m}{s ^ { 2 }} $.

**Expert Answer**

**Step 1**

**Given values**

\[ h = 1.0 m\]

\[\theta = 20 ^ { \circ } \]

\[ g = 9.81 \dfrac{ m } { s ^ { 2 } } \]

**Step 2**

When the **piano starts moving down the ramp**, the**Â gravitational acceleration**Â is:

\[a = g \sin \theta \]

If we **substitute the values into the above equation,**Â we get the desired **acceleration value**:

\[a = ( 9.81 \dfrac {m}{ s ^{2}})( \sin ( 20 ^ Â { \circ } ))\]

\[a = ( 9.81 \dfrac{ m }{ s ^ { 2 }} )( 0.34202 )\]

\[a = 3.35 \dfrac{m}{s ^ { 2 }} \]

**Length of the ramp is given**Â as:

\[\sin \theta = \dfrac {h}{\Delta x}\]

\[\Delta x = \dfrac{h}{\sin\theta}\]

\[\Delta x = \dfrac{1.0}{\sin (20^{\circ})}\]

\[\Delta x = \dfrac{1.0}{0.34202}\]

\[\Delta x = 2.92m\]

So the **time for the piano to reach the ground**Â is:

\[t = \sqrt {\dfrac{\Delta x}{a}}\]

\[t = \sqrt {\dfrac{2.92m}{3.35 \dfrac{m}{s^{2}}}}\]

\[t = 1.32 s\]

The **time**Â is $ 1.32s $.

**Numerical Result**

The **time it takes the workers to reach the piano before it reaches the bottom**Â of the ramp is $ 1.32 s$.

**Example**

**The piano was pushed to the top of the ramp in the back of the moving van. The workers think it’s safe, but as they leave, it starts rolling down the ramp. If the back of the truck is $2.0\: m$ above the ground and the ramp is inclined $ 30^{\circ}$, how much time will the workers take to get to the piano before it reaches the bottom of the ramp?**

**Solution**

**Step 1**

**Given values**

\[ h = 2.0 m\]

\[\theta = 30^ {\circ} \]

\[g = 9.81 \dfrac{m}{s^{2}} \]

**Step 2**

When the **piano starts moving down the ramp**, the**Â gravitational acceleration**Â is:

\[a = g \sin \theta \]

If we **substitute the values into the above equation,**Â we get the desired **acceleration value**:

\[a = (9.81 \dfrac{m}{s^{2}} )(\sin(30^ {\circ}))\]

\[a = (9.81 \dfrac{m}{s^{2}} )(0.5)\]

\[a = 19.62 \dfrac{m}{s^{2}} \]

**Length of the ramp is given**Â as:

\[\sin \theta = \dfrac{h}{\Delta x} \]

\[\Delta x = \dfrac{h}{\sin \theta } \]

\[\Delta x = \dfrac{2.0}{\sin (30^{\circ})}\]

\[\Delta x = \dfrac{1.0}{0.5}\]

\[\Delta x = 4m\]

So the **time for the piano to reach the ground**Â is:

\[t = \sqrt {\dfrac{\Delta x}{a}}\]

\[t = \sqrt {\dfrac{4m}{19.62 \dfrac{m}{s^{2}}}} \]

\[t = 0.203 s\]

The **time**Â is $ 0.203s $.