A plane, diving with constant speed at an angle of 53.0^{\circ}

To solve the given problem, let's break it down into parts:
a) Finding the speed of the plane:
The vertical motion of the plane is independent of its horizontal motion. Since the plane is diving at an angle of 53.0 degrees with the vertical and the time of flight for the projectile is given as 5.00 s, we can use the equation for vertical displacement:
$y=v_{0y}t+\frac{1}{2}g{t}^2$
where $y$ is the vertical displacement, $v_{0y}$ is the vertical component of initial velocity, $t$ is the time, and $g$ is the acceleration due to gravity. Here, $y=-730\text{m}$ (taking downward direction as negative), $t=5.00\text{s}$, and $g=9.8{\text{m/s}}^2$.
Since the plane is diving at an angle of 53.0 degrees with the vertical, the initial vertical velocity can be expressed as:
$v_{0y}=v_p\sin \theta$
where $v_p$ is the speed of the plane and $\theta$ is the angle with the vertical. Here, $\theta ={53.0}^{\circ }$. By substituting the given values into the equations, we can solve for $v_p$.
b) Finding the horizontal distance traveled by the projectile:
The horizontal motion of the projectile is unaffected by the vertical motion of the plane. We can use the equation for horizontal distance:
$x=v_{0x}t$
where $x$ is the horizontal distance, $v_{0x}$ is the horizontal component of initial velocity, and $t$ is the time of flight. Since there is no acceleration in the horizontal direction, $v_{0x}$ remains constant throughout the motion. By substituting the given values into the equation, we can calculate $x$.
c) Finding the horizontal component of velocity just before striking the ground:
The horizontal component of velocity remains constant throughout the motion. Thus, the horizontal component of velocity just before striking the ground is equal to the horizontal component of initial velocity.
d) Finding the vertical component of velocity just before striking the ground:
The vertical component of velocity changes due to the effect of gravity. We can use the equation for vertical velocity:
$v_y=v_{0y}+gt$
where $v_y$ is the vertical component of velocity, $v_{0y}$ is the initial vertical component of velocity, $g$ is the acceleration due to gravity, and $t$ is the time of flight. By substituting the given values into the equation, we can calculate $v_y$.

Result:
a) The speed of the plane is approximately 44.4 m/s.
b) The projectile travels approximately 150 m horizontally during its flight.
c) The horizontal component of the velocity just before striking the ground is approximately 44.4 m/s.
d) The vertical component of the velocity just before striking the ground is approximately -6.02 m/s.
Solution:
a) To find the speed of the plane, we can use the vertical motion of the projectile. The vertical displacement of the projectile is given by the equation:
$y=v_{i_y}t+\frac{1}{2}g{t}^2$
where $y$ is the vertical displacement (730 m), $v_{i_y}$ is the initial vertical velocity of the projectile, $t$ is the time of flight (5.00 s), and $g$ is the acceleration due to gravity (-9.8 m/s${}^2$).
Since the plane is diving at an angle of 53.0 degrees with the vertical, the initial vertical velocity of the projectile can be determined using trigonometry:
$v_{i_y}=v\sin (\theta )$
where $v$ is the speed of the plane and $\theta$ is the angle of dive (53.0 degrees).
Substituting this expression into the first equation and solving for $v$, we get:
$y=v\sin (\theta )t+\frac{1}{2}g{t}^2$
Simplifying further, we have:
$v=\frac{y-\frac{1}{2}g{t}^2}{\sin (\theta )t}$
Substituting the given values, we get:
$v=\frac{730\text{m}-\frac{1}{2}\times 9.8{\text{m/s}}^2\times (5.00\text{s}{)}^2}{\sin ({53.0}^{\circ })\times 5.00\text{s}}$
Calculating the value, we find:
$v\approx 44.4\text{m/s}$
Therefore, the speed of the plane is approximately 44.4 m/s.
b) The horizontal distance traveled by the projectile during its flight can be calculated using the horizontal motion equation:
$x=v_{i_x}t$
where $x$ is the horizontal distance traveled, $v_{i_x}$ is the initial horizontal velocity of the projectile, and $t$ is the time of flight (5.00 s).
The initial horizontal velocity $v_{i_x}$ can be determined using trigonometry:
$v_{i_x}=v\cos (\theta )$
Substituting the given values, we have:
$x=v\cos (\theta )t$
Substituting the calculated value of $v$ and the given values of $\theta$ and $t$, we get:
$x=44.4\text{m/s}\times \cos ({53.0}^{\circ })\times 5.00\text{s}$
Calculating the value, we find:
$x\approx 150\text{m}$
Therefore, the projectile travels approximately 150 m horizontally during its flight.
c) The horizontal component of the velocity just before striking the ground remains constant throughout the motion and is equal to the initial horizontal velocity, $v_{i_x}$. Using the previously calculated value of $v_{i_x}$, we have:
$v_{i_x}\approx 44.4\text{m/s}$
Therefore, the horizontal component of the velocity just before striking the ground is approximately 44.4 m/s.
d) The vertical component of the velocity just before striking the ground can be determined using the equation:
$v_y=v_{i_y}+gt$
where $v_y$ is the vertical component of the velocity, $v_{i_y}$ is the initial vertical velocity of the projectile, $g$ is the acceleration due to gravity (-9.8 m/s²), and $t$ is the time of flight (5.00 s).
The initial vertical velocity $v_{i_y}$ can be calculated using trigonometry:
$v_{i_y}=v\sin (\theta )$
Substituting the given values, we have:
$v_y=v\sin (\theta )+gt$
Substituting the calculated value of $v$, the given values of $\theta$ and $t$, and the acceleration due to gravity, we get:
$v_y=44.4\text{m/s}\times \sin ({53.0}^{\circ })+(-9.8\text{m/s²})\times 5.00\text{s}$
Calculating the value, we find:
$v_y\approx -6.02\text{m/s}$
Therefore, the vertical component of the velocity just before striking the ground is approximately -6.02 m/s. The negative sign indicates that the velocity is directed downward.

Step 1:
a) To solve for the speed of the plane, we can consider the vertical motion of the projectile. The time it takes for the projectile to hit the ground is given as $t=5.00$ s, and the initial vertical displacement is $y_0=730$ m.
We can use the equation of motion for vertical motion under constant acceleration: $y=y_0+v_{0y}t-\frac{1}{2}g{t}^2$, where $v_{0y}$ is the initial vertical component of velocity and $g$ is the acceleration due to gravity.
Since the plane is diving at an angle of ${53.0}^{\circ }$ with the vertical, the initial vertical component of velocity can be found using $v_{0y}=v_0\sin (\theta )$, where $v_0$ is the speed of the plane and $\theta ={53.0}^{\circ }$.
Substituting the given values into the equation, we have $0=730+v_0\sin ({53.0}^{\circ })·5.00-\frac{1}{2}·9.8·(5.00{)}^2$.
Solving this equation for $v_0$, we find:
$v_0=\frac{730}{5.00\sin ({53.0}^{\circ })}+\frac{1}{2}·9.8·5.00\approx 56.0\text{m/s}.$
Therefore, the speed of the plane is approximately $56.0$ m/s.
Step 2:
b) The horizontal distance traveled by the projectile can be determined using the equation $x=v_{0x}t$, where $v_{0x}$ is the initial horizontal component of velocity. Since there is no horizontal acceleration, the horizontal component of velocity remains constant throughout the motion.
The initial horizontal component of velocity can be calculated using $v_{0x}=v_0\cos (\theta )$, where $\theta ={53.0}^{\circ }$ and $v_0$ is the speed of the plane.
Substituting the given values into the equation, we have $x=v_0\cos ({53.0}^{\circ })·5.00$.
Evaluating this expression using the previously determined value for $v_0$, we find:
$x=56.0\cos ({53.0}^{\circ })·5.00\approx 182\text{m}.$
Therefore, the projectile travels approximately 182 m horizontally during its flight.
Step 3:
c) The horizontal component of velocity just before striking the ground remains unchanged. Therefore, the horizontal component of velocity is the same as $v_{0x}$, which we previously calculated as $v_{0x}=v_0\cos (\theta )$.
Substituting the known values, we have:
$v_{0x}=56.0\cos ({53.0}^{\circ })\approx 33.6\text{m/s}.$
Therefore, the horizontal component of velocity just before striking the ground is approximately 33.6 m/s.
Step 4:
d) The vertical component of velocity just before striking the ground can be determined using the equation $v_y=v_{0y}-gt$, where $v_{0y}$ is the initial vertical component of velocity and $t=5.00$ s.
Substituting the known values, we have:
$v_y=v_0\sin ({53.0}^{\circ })-9.8·5.00\approx 0\text{m/s}.$
Therefore, the vertical component of velocity just before striking the ground is approximately 0 m/s.