The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 10.0m/s. (a) What is the magnitude of the velocity of the projectile 1.00 sbefore it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.00 safter it achieves its maximum height? If we take x=0and y=0to be at the point of maximum height and positiveto be in the direction of the velocity there, what are the (c) x-coordinate and (d) y-coordinate of the projectile 1.00 sbefore it reaches its maximum height and the (e)coordinate and (f)coordinate 1.00 safter it reaches its maximum height?

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 .