Q54P - Fundamentals Of Physics

Question

A stone is dropped into a river from a bridge 43.9m above the water. Another stone is thrown vertically down 1.00s after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on the graph for each stone, taking zero time as the instant the first stone is released.

Step-by-Step Solution

Answer

a) Initial speed for second stone is 12.3 m/s

b) Graph for velocity vs time is plotted

Step 1: Given Data

The height of bridge is, s = 43.9

Acceleration acting on both stones is equal to gravitational acceleration, $g=9.8 m/s^2$

Step 2: Understanding the kinematic equations of motion

Kinematic equations of motion describe the relationship between displacement, velocity or acceleration of a particle.

From kinematic equations of motion, the equation for displacement is written as

$s=v_{0}t+\frac{1}{2}a{t}^2$                                                                                                                                             

(i)

Step 3: (a) Determining the initial speed of second stone

The initial velocity of the stone is, $v_0=0 m/s$. Putting the given values in the equation (i)

$$\begin{gathered} 43.9m=0+\frac{1}{2}\times 9.8m/s^2\times t^2 \\ t^2=\frac{43.9\times 2}{9.8} \\ t=\sqrt{\frac{87.8}{9.8}} \\ =2.99 s \end{gathered}$$