91Ó°ÊÓ

The given data can be listed below,

• The acceleration of the rocket is,$a=1.90m/s^2$
• The angle of inclination is,$\mathrm{θ}=35°$

A projectile's horizontal range is the length of the horizontal plane it would traverse before returning to its initial vertical location.

The initial height of the projectile launched is given by,

$y_0=d\sin \mathrm{θ}$

Here, d is the inclined distance.

Substitute all the values in the above,

$$\begin{gathered} y_0=\left(200m\right)\sin 35° \\ =114.7m \end{gathered}$$

The vertical component of initial velocity is given by,

$v_y=v\sin \mathrm{θ}$

Here, v is the velocity after a displacement d.

Substitute all the values in the above,

$$\begin{gathered} v_{0y}=v\sin 35° \\ =\sqrt{2ad}\sin 35° \\ =\sqrt{2\left(1.90m/s^2\right)\left(200m\right)}\sin 35° \\ =15.81m/s \end{gathered}$$

The equation for vertical motion of the projectile is given by,

$v_y^2=v_{0y}^2-2g\left(y-y_0\right)$

Here,$v_{0y}$ is initial vertical velocity of the projectile, g is the acceleration due to gravity, y is the maximum height of the projectile.

At maximum height, vertical velocity of the projectile is zero so the maximum height is given by,

$$\begin{gathered} 0=v_{0y}^2-2g\left(y_{max}-y_0\right) \\ {y}_{max}=y_0+\frac{v_{0y}^2}{2g} \end{gathered}$$

Substitute all the value in the above,

$$\begin{gathered} y_{max}=114.7m+\frac{\left(15.81m/s^2\right)}{2\left(9.8m/s^2\right)} \\ =127.45m \\ ≈128m \end{gathered}$$

Thus, the maximumheight of the rocket above the ground is 128 m .

The time of flight of the rocket is given by,

$t=-\frac{1}{g}\left(-v_{0y}Â\pm \sqrt{v_{0y}^2+2g{y}_0}\right)$

Here, $v_{0y}$is initial vertical velocity of the projectile, g is the acceleration due to gravity,$y_0$ is the maximum height of the projectile.

Substitute all the values in the above,

$$\begin{gathered} t=-\frac{1}{9.8m/s^2}\left[-15.81m/sÂ\pm \sqrt{{\left(15.81m/s\right)}^2+2\left(9.8m/s^2\right)\left(114.7m\right)}\right] \\ =1.61sÂ\pm 5.10s \end{gathered}$$

The positive value of time is taken to find the horizontal range which can be given by,

$x_{max}=d\cos \mathrm{θ}+t\sqrt{2ad}\cos \mathrm{θ}$

Here, d is the inclined distance, a is the acceleration, and t is the time of flight.

Substitute all the values in the above,

$$\begin{gathered} x_{max}=\left(200s\right)\cos 35°+\sqrt{2\left(1.90m/s^2\right)\left(200m\right)}\cos 35° \\ =315m \end{gathered}$$

Thus, the maximum horizontal range of the rocket is 315 m