Question

Please help I don’t understand this is from a PowerPoint from my self paced class.

Answer 1

Given data:

* The height of bridge above the water is 120 m.

* The initial horizontal speed of the stone is 25 m/s.

Solution:

(a). By the kinematics equation, the vertical motion of the stone is,

`H=u_yt+(1)/(2)gt^2`

where u_y is the vertical initial velocity of the stone, g is the acceleration due to gravity, t is the time taken, and H is the height of the bridge from the water,

The vertical initial velocity of the stone is zero.

Substituting the known values,

`\begin{gathered} 120=0+(1)/(2)*9.8* t^2 \\ 120=4.9* t^2 \\ t^2=(120)/(4.9) \\ t^2=24.49 \\ t=4.95\text{ s} \end{gathered}`

Thus, the time taken by the stone to strike the water is 4.95 seconds.

(b).The horizontal range of the stone is,

`x=u_xt+(1)/(2)at^2`

where u_x is the initial horizontal velocity and a is the acceleration along the horizontal direction, t is thetime taken, and x is the horizonatl range,

The acceleration of the stone in the horizontal direction is zero.

Substituting the known values,

`\begin{gathered} x=25*4.95 \\ x=123.75\text{ m} \end{gathered}`

As there is no acceleration in the horizontal direction, thus, the final value of velocity in the horizontal direction is same as the initial horizontal direction.

The values of the horizontal motion variables are,

`\begin{gathered} x_o=0 \\ x=123.75\text{ m} \\ v_o=25\text{ m/s} \\ v=25\text{ m/s} \\ a=0ms^(-2) \\ t=4.95\text{ s} \end{gathered}`

(c). By the kinematics equation, the final vertical velocity of the stone is,

`v-u=gt`

where v is the final vertical velocity, u is the initial vertical velocity, g is the acceleration due to gravity and t is the time taken,

Substituting the known values,

`\begin{gathered} v-0=9.8*4.95 \\ v=48.51\text{ m/s} \end{gathered}`

Let the distance of stone is measured in the downward direction, then,

The values of variables for the vertical motion of stone are,

`\begin{gathered} y_o=0\text{ m} \\ y=120\text{ m} \\ v_(oy)=0\text{ m/s} \\ v_y=48.51\text{ m/s} \\ a=9.8ms^(-2) \\ t=4.95\text{ s} \end{gathered}`