Question

An artillery shell is fired at an angle of 62.7

above the horizontal ground with an initial
speed of 1520 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell,
neglecting air resistance.
Answer in units of min ...?

Answer 1

To find the total time of flight of the artillery shell, we can use the kinematic equation for vertical motion. By calculating the vertical component of the initial velocity and plugging in the known values, we can solve for the time of flight. Finally, we can convert the time to minutes by dividing by 60.

Step-by-step explanation:

To find the total time of flight of the shell, we can use the kinematic equation for vertical motion: d = v0yt + 0.5gt2. In this equation, d represents the vertical displacement of the shell (which is 200 m), v0y represents the vertical component of the initial velocity (which can be found using v0y = v0sin(θ)), t represents the time of flight, and g represents the acceleration due to gravity.

Since the shell is fired at an angle of 62.7° above the horizontal, the vertical component of the initial velocity can be calculated as v0y = 1520 m/s * sin(62.7°). Plugging in the known values, we can solve for t.

Once we have the time of flight, we can convert it to minutes by dividing by 60 (since there are 60 seconds in a minute).

Answer 2

This is a motion problem. For bodies in a projectile motion, several equations have already been derived and simplified to account for the behavior of these bodies.
For the time of flight, the equation is

t = (2*v*sin(A))/g

where
v = initial velocity of the body
A = angle of the initial velocity
g = acceleration due to gravity (9.81 m/s^2)
t = time of flight

this formula can also be used if the air resistance can be neglected.

therefore, substituting the values, we get

t = 275.65 s or 4.59 min