91Ó°ÊÓ

1) Initial velocity of the projectile is${\mathrm{V}}_0=30\mathrm{m}/\mathrm{s}$

2) The launch angle of the projectile is,${\mathrm{θ}}_0=60°$

The motion of an object launched vertically into the air and travelling under the effect of gravity is known as projectile motion. The motion of a projectile thrown at an angle and under gravitational acceleration can be described using the kinematic equations. Using the kinematic equations, the magnitude and angle of velocity at different times can be found.

Formula:

$$\begin{gathered} V_x=V_0\cos {\mathrm{θ}}_0 \\ {V}_y=V_0\sin {\mathrm{θ}}_0-gt \\ V=\sqrt{V_x^2+V_y^2} \\ \mathrm{θ}={\tan }^{-1}\left(\frac{V_y}{V_x}\right) \end{gathered}$$

There is no acceleration in the horizontal direction. Therefore, the velocity of the projectile in the horizontal direction will not change with time.Therefore, we can write,

$$\begin{gathered} V_x=V_0\cos {\mathrm{θ}}_0 \\ =15m/s \end{gathered}$$

There is a gravitational acceleration acting in the downward direction. Therefore, the vertical component of the velocity will change with respect to time. Hence, the y component of the velocity of projectile is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{y}}={\mathrm{V}}_0{\mathrm{²õ¾\pm ²Ôθ}}_0-\mathrm{gt} \\ =\left(30\mathrm{m}/\mathrm{s}\right)\sin \left(60°\right)-\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+\left(2.0\mathrm{s}\right) \\ =6.38\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the magnitude of velocity after $2.0s$is given by,

$$\begin{gathered} \mathrm{V}=\sqrt{{\mathrm{V}}_{\mathrm{x}}^2+{\mathrm{V}}_{\mathrm{y}}^2} \\ =\sqrt{{\left(15.0\mathrm{m}/\mathrm{s}\right)}^2+{\left(6.38\mathrm{m}/\mathrm{s}\right)}^2} \\ =16.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the magnitude of the projectile velocity after $2.0\mathrm{s}$is $16.3\mathrm{m}/\mathrm{s}$.

The angle of velocity is given by the equation.

$\mathrm{θ}={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$

Substitute the value of velocity components from part a)

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{V_y}{V_x}\right) \\ ={\tan }^{-1}\left(\frac{6.38m/s}{15.0m/s}\right) \\ =23° \end{gathered}$$

Therefore, the angle of the projectile velocity $2.0s$after the launch is $23°$.

From part b), the angle made by the velocity after $2.0s$is positive. It implies that the velocity is directed above the horizontal.

Following the procedure of part a), for$t=5.0s$, we get

$$\begin{gathered} {\mathrm{V}}_{\mathrm{x}}={\mathrm{V}}_0{\mathrm{³¦´Ç²õθ}}_0 \\ =15\mathrm{m}/\mathrm{s} \\ {\mathrm{V}}_{\mathrm{y}}={\mathrm{V}}_0{\mathrm{²õ¾\pm ²Ôθ}}_0-\mathrm{gt} \\ =\left(30\mathrm{m}/\mathrm{s}\right)\sin \left(60°\right)-\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)×\left(5.0\mathrm{s}\right) \\ =-23.02\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the magnitude of velocity is,

$$\begin{gathered} \mathrm{V}=\sqrt{{\mathrm{V}}_{\mathrm{x}}^2+{\mathrm{V}}_{\mathrm{x}}^2} \\ =\sqrt{{\left(15.0\mathrm{m}/\mathrm{s}\right)}^2+{\left(-23.02\mathrm{m}/\mathrm{s}\right)}^2} \\ =27.47\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the magnitude of the projectile velocity after $5.0s$ is $27.47\mathrm{m}/\mathrm{s}$.

The angle of the velocity at $t=5.0s$will be,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{V_y}{V_x}\right) \\ ={\tan }^{-1}\left(\frac{-23.02m/s}{15.0m/s}\right) \\ =-57° \end{gathered}$$

Therefore, the angle of the projectile velocity$5.0s$ after the launch is$-57°$.

From part e), the angle made by the velocity after $5.0\mathrm{s}$is negative. It implies that the velocity is directed below the horizontal.