A shell is fired from the ground with an initial speed of 1.68 103 m/s at an initial angle of 37° to the horizontal.

(a) Neglecting air resistance, find the shell's horizontal range.
m
(b) Find the amount of time the shell is in motion.
s

time in air:

vertical speed initial: 1.68E3 sin 37
hf=hi+Vi*t-1/2 g t^2
0=0+1.68E3 sin 37 *t-4.8t^2
solve for time of flight, t.

horizontal range:
distance= 1.68E3*cos37*t

To find the horizontal range of the shell, we need to determine the time of flight and horizontal displacement.

(a) To find the time of flight, we can use the vertical motion equation:
vf = vi + at

Since the shell is fired from the ground, its initial vertical velocity is 0 and the only force acting on it is gravity, so the acceleration is -9.8 m/s² (taking downward as the negative direction).

At the maximum height, the vertical velocity will also be 0 (v = 0). We can use this information to find the time it takes for the shell to reach its peak:

0 = viy + ay * t
0 = (1.68 * 10^3 * sin(37°)) + (-9.8) * t

Solving for t gives us:
t = (1.68 * 10^3 * sin(37°)) / 9.8

Now, we can double this time to find the total time of flight (as the shell will spend the same amount of time on the way down):
t_total = 2 * t

Now, to find the horizontal displacement, we can use the horizontal motion equation:
x = v * t

Since the initial velocity only has a horizontal component, we can multiply it by the time of flight to find the range:
range = (1.68 * 10^3 * cos(37°)) * t_total

Plugging in the values:
range = (1.68 * 10^3 * cos(37°)) * (2 * [(1.68 * 10^3 * sin(37°)) / 9.8])

Simplifying this equation will give us the horizontal range of the shell.

(b) To find the amount of time the shell is in motion, we already found it in part (a). The total time of flight is:
t_total = 2 * t

Substituting the values we have:
t_total = 2 * [(1.68 * 10^3 * sin(37°)) / 9.8]

Once simplified, this equation will give us the time the shell is in motion.

To find the shell's horizontal range, we need to use the equation for projectile motion:

Horizontal Range (R) = (Initial velocity * time of flight) * cosine(angle)

Given:
Initial velocity (v₀) = 1.68 * 10^3 m/s
Angle (θ) = 37° (converted to radians)

To find the time of flight, we can use the equation for the vertical component of the shell's motion:

Vertical Displacement (y) = (initial vertical velocity * time) + (1/2 * acceleration due to gravity * time^2)

Since the shell is fired from the ground, the initial vertical velocity (v₀y) is equal to zero. The vertical displacement (y) is also equal to zero at the maximum height. We can use these values to find the time of flight.

(a) Finding the shell's horizontal range:

R = (v₀ * t) * cos(θ)

R = (1.68 * 10^3 m/s * t) * cos(37°)

(b) Finding the time of flight:

At the maximum height of the projectile's trajectory, the vertical displacement is zero. Using this information, we can determine the time it takes for the shell to reach its highest point.

0 = (0 * t) + (1/2 * (-9.8 m/s^2) * t^2)

Simplifying the equation:

0 = -4.9 * t^2

Solving for t:

t^2 = 0

t = 0

Since the time of flight cannot be zero, this result indicates that the shell never reaches its maximum height. Instead, it reaches its maximum vertical displacement at half the time of flight. Thus, the total time of flight is:

Time of flight (T) = 2 * t

T = 2 * 0

T = 0 s

Therefore, the shell is in motion for 0 seconds.