Q49E - University Physics with Modern Physics

Question

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 6.00 s after it was thrown. What is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? Ignore air resistance.

Step-by-Step Solution

Answer

Speed of the rock just before it reaches the water is 37.62 m/s.

Step 1: Identification of given values

Displacement (s) =-28 m

Time = 6 s

Step 2: calculation for the initial velocity

After 6 seconds displacement is zero

We have equation of motion$s=ut+\frac{1}{2}a{t}^2$

Acceleration, a = -9.81 m/s²  

Time, t = 6 s      

After 6 sceond Displacement, s = 0

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 0=u\times 6+\frac{1}{2}\times -9.81\times 6^2 \\ u=29.43\mathrm{m}/\mathrm{s} \end{gathered}$$ 

Step 3: calculation for speed

As we know that$v^2=u^2+2as$ 

Here s = -28 m

$$\begin{gathered} v^2=u^2+2as \\ {v}^2=29.{43}^2+2\times \left(-9.81\right)\times \left(-28\right) \\ v=37.62\mathrm{m}/\mathrm{s} \end{gathered}$$