91Ó°ÊÓ

The magnitude of velocity at maximum height,$V_f=10.0\mathrm{m}/\mathrm{s}$ .

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s having horizontal directional acceleration is zero and its vertical directional acceleration is the free-fall acceleration . (Upward is taken to be a positive direction.)

We can find the velocity of the projectile at a given point by using components of the initial velocity. We can find the coordinates of projectiles by using a kinematic equation.

Formulae:

The first kinematic equation of motion,$V_f=V_i+at$ (i)

The second kinematic equation of motion,$d=d_0+V_{i}t+\left(\frac{1}{2}\right)a{t}^2$ (ii)

We know that at maximum height, the vertical velocity component becomes $V_{fy}=0$and there is only a horizontal velocity component,role="math" localid="1657024064243" $V_{fx}=10.0\mathrm{m}/\mathrm{s}$ .

Before reaching to maximum height initial velocity is $V_{ix}=10.0\mathrm{m}/\mathrm{s}$, we can find initial vertical velocity at 1.00s by using final velocity$V_{fy}=0$at maximum height.

Using equation (i) and the given values, we get the initial velocity as follows:

$$\begin{gathered} 0=V_{iy}+\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.00\mathrm{s}\right) \\ {V}_{iy}=9.8\mathrm{m}/\mathrm{s} \end{gathered}$$

We can find resultant velocity by using horizontal and vertical components,

$$\begin{gathered} V_{f1}=\sqrt{{\left(10.0\mathrm{m}/\mathrm{s}\right)}^2+{\left(9.8\mathrm{m}/\mathrm{s}\right)}^2} \\ =14\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the magnitude of velocity of projectile at 1.00s before it achieves maximum height is 14m/s.

Here initial vertical velocity$V_{iy}=0$at maximum height, we find$V_{iy}$after 1.00 s. Hence, the final velocity using equation (i) and the given values is given as:

$$\begin{gathered} V_{fy}=0+{\left(-9.8\mathrm{m}/\mathrm{s}\right)}^2\left(1.00s\right) \\ =-9.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Using resultant velocity equation we get, the velocity after reaching maximum height as follows:

$$\begin{gathered} V_{f2}=\sqrt{{\left(10.0\mathrm{m}/\mathrm{s}\right)}^2+{\left(-9.8\mathrm{m}/\mathrm{s}\right)}^2} \\ =14\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the velocity after reaching maximum height is 14m/s .

We have to take x=0 and y=0 at maximum height.

Using equation (ii) and the given, we get the x-coordinateas follows:

Since $a_x=0$along horizontal,

$x=x_1+V_{ix}t=0$

Substitute the values in above equation,

$$\begin{gathered} 0=x_i+\left(10.0\mathrm{m}/\mathrm{s}\right)\left(1.00\mathrm{s}\right) \\ {x}_i=-10\mathrm{m} \end{gathered}$$

Hence, the value of x-coordinate is -10m .

The y-coordinate can be given using equation (ii) and the given values, we get

$$\begin{gathered} 0=y_1+{\left(9.8\mathrm{m}/\mathrm{s}\right)}^2\left(1.00\mathrm{s}\right)+\left(0.5\right)\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(1.00s\right)}^2 \\ {y}_1=-9.8\mathrm{m}+4.9\mathrm{m} \\ =-4.9\mathrm{m} \end{gathered}$$

Hence, the value of y-coordinate is -4.9m.

Using equation (ii) of distance and the given values for xcoordinateafter 1.00 s we get the x-coordinate as follows:

$$\begin{gathered} x_f=0+\left(10.0\mathrm{m}/\mathrm{s}\right)\left(1.00\mathrm{s}\right)+0 \\ =10m \end{gathered}$$

Hence, the x-coordinate of projectile is 10 m.

Using equation (ii) of distance and the given values for y-coordinateafter 1.00s we get the y-coordinate as follows:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{f}}=0+\left(0\right)\left(1.00\mathrm{s}\right)+\left(0.5\right)\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(1.00\mathrm{s}\right)}^2 \\ =-4.9\mathrm{m} \end{gathered}$$

Hence, the y-coordinate of projectile is -4.9m .