Question

A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out bridge (Fig. P4.76). The quick stop causes a number of melons to fly off the truck. One melon leaves the hood of the truck with an initial speed $v_i=10.0\text{m/s}$in the horizontal direction. A cross section of the bank has the shape of the bottom half of a parabola, with its vertex at the initial location of the projected watermelon and with the equation $y^2=16x$where and are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank?

Short answer

The required impact occurs at $y=17.34\text{m}$down and $x=18.8\text{m}$beyond the edge.

Step-by-step solution

Define hhorizontals projection

A projectile's horizontal projection Assume a body is hurled horizontally with velocity u from a location$\mathrm{O}.$ .

The height of point$\mathrm{O}.$above the earth is h

The body has two independent motions going on at the same time.

Find the path of the horizontal projection

The equation is$y^2=16x$ .

The horizontal displacement of the projectile for horizontallyprojected bodies,

$x=v_{i}t$

Here, ${\text{v}}_{\text{i}}$is the initial speed, and t is the time.

For horizontallyprojected bodies, the projectile's horizontal displacement,

$y=-\frac{1}{2}g{t}^2$

Here, g is the acceleration due to gravity.

The equation for $x=v_{i}t$for t ,

$$\begin{gathered} x=v_{i}t \\ t=\frac{x}{v_i} \end{gathered}$$

Now, substitute $\frac{\text{x}}{{\text{v}}_{\text{i}}}$for in the equation for the vertical displacement of the projectile as follows:

$$\begin{gathered} y=-\frac{1}{2}g{t}^2 \\ =-\frac{1}{2}g{\left(\frac{x}{v_i}\right)}^2{y}^2 \\ ={\left[-\frac{1}{2}g{\left(\frac{x}{v_i}\right)}^2\right]}^2 \end{gathered}$$

By equating the equations,$y^2=16x$and $y^2={\left[-\frac{1}{2}g{\left(\frac{x}{v_i}\right)}^2\right]}^2$

$y^2={\left[-\frac{1}{2}g{\left(\frac{x}{v_i}\right)}^2\right]}^2$

$x={\left(\frac{64v_i^4}{g^2}\right)}^{\frac{1}{3}}$

Substitute$10.0\text{m/s for}{v}_i\text{and}9.80{\text{m/s}}^{\text{2}}\text{for}g\text{.}$

$$\begin{gathered} x=4{\left(\frac{v_i^4}{g^2}\right)}^{\frac{1}{3}} \\ ={\left[\frac{\left(64\right){(10.0\text{m/s})}^4}{{\left(9.80{\text{m/s}}^{\text{2}}\right)}^2}\right]}^{\frac{1}{3}} \\ =18.8\text{m} \end{gathered}$$

Now find the vertical displacement

$$\begin{gathered} y^2=16x \\ y=\sqrt{16x} \end{gathered}$$

Substitute $18.8\text{m}$for x.

$$\begin{gathered} y=\sqrt{16x} \\ =\sqrt{16(18.8\text{m})} \\ =17.34\text{m} \end{gathered}$$

Therefore, the required impact occurs at $y=17.34\text{m}$down and $x=18.8\text{m}$beyond the edge.