Question

The height$\mathrm{y}$ and the distance $\mathrm{x}$along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $\mathrm{y}=(8\mathrm{t}-5{\mathrm{t}}^2)\mathrm{m}\mathrm{and}\mathrm{x}=6\mathrm{t}\mathrm{m},$ where t is in seconds. The velocity with which the projectile is projected at $\mathrm{t}=0$ is.?

Answer 1

Step 1: Given data:

The height$\mathrm{y}$ of a projectile on a certain planet is given by =$\mathrm{y}=(8\mathrm{t}-5{\mathrm{t}}^2)\mathrm{m}$

The distance $\mathrm{x}$along the horizontal plane is given by =$\mathrm{x}=6\mathrm{t}\mathrm{m}$

(Where$\mathrm{t}$is in seconds)

Step 2: Calculating Velocity in $\mathrm{x}\mathrm{and}\mathrm{y}$-direction:
We know velocity is defined as the rate of change of displacement,

Hence, the Velocity in the $\mathrm{x}-$direction can be given as:

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ =\frac{\mathrm{d}}{\mathrm{dt}}6\mathrm{t} \\ =6\mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity in the $\mathrm{y}-$direction can be given as:

$$\begin{gathered} {\mathrm{u}}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}} \\ =\frac{\mathrm{d}}{\mathrm{dt}}(8\mathrm{t}-5{\mathrm{t}}^2) \\ =8-10\mathrm{t} \end{gathered}$$
At time $\mathrm{t}=0$,

${\mathrm{u}}_{\mathrm{y}}=8\mathrm{m}/\mathrm{s}$

Step 3: Calculating the Velocity of projectile at $\mathrm{t}=0\mathrm{s}$:
The resultant velocity of the projectile at $\mathrm{t}=0\mathrm{s}$ can be calculated as:

$$\begin{gathered} \mathrm{u}=\sqrt{{{\mathrm{u}}_{\mathrm{x}}}^2+{{\mathrm{u}}_{\mathrm{y}}}^2} \\ =\sqrt{6^2+8^2}\mathrm{m}/\mathrm{s} \\ =10\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the velocity with which the projectile is projected at $\mathrm{t}=0$ is $10\mathrm{m}/\mathrm{s}$.