A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an initial angle of 43.0◦ to the horizontal.
The acceleration of gravity is 9.81 m/s2 .
a) Neglecting air resistance, find the shell’s horizontal range.
Answer in units of m.
b) How long is the shell in motion? Answer in units of s.
a) Oh, the shell wants to go on a little field trip! Let's calculate its horizontal range. Now, we know that the initial velocity of the shell has two components: one in the horizontal direction and one in the vertical direction. Since we're interested in the horizontal range, we only need to focus on the horizontal component of the velocity. So, using a little trigonometry magic, we find that the initial horizontal velocity is 1.60 × 103 m/s * cos(43.0°). Once we have this, we can calculate the time it takes for the shell to hit the ground using the formula t = 2 * (v_0_x) / g, where v_0_x is the initial horizontal velocity and g is the acceleration due to gravity. Then, we can find the horizontal range using the formula R = v_0_x * t. Crunching the numbers, the shell's horizontal range is... *drum roll please*... Answer in units of m!
b) Now, let's determine how long the shell is in motion. Since we already calculated the time it takes for the shell to hit the ground in part a), we can use that value as our answer. So, the shell is in motion for... *another drum roll*... units of s! Enjoy the show!
To find the shell's horizontal range, we need to calculate the time of flight.
a) The time of flight (T) can be calculated using the formula:
T = 2 * (V₀ * sinθ) / g
Where:
V₀ = initial speed = 1.60 × 10³ m/s
θ = initial angle = 43.0°
g = acceleration due to gravity = 9.81 m/s²
Substituting the given values:
T = 2 * (1.60 × 10³ * sin(43.0°)) / 9.81
Calculating the value:
T ≈ 40.91 s
Now, to find the horizontal range (R), we can use the formula:
R = V₀ * cosθ * T
Substituting the given values:
R = (1.60 × 10³ * cos(43.0°)) * 40.91
Calculating the value:
R ≈ 2.06 × 10⁵ m
Therefore, the shell's horizontal range is approximately 2.06 × 10⁵ m.
b) Since we already know the time of flight (T) from part a, the shell is in motion for approximately 40.91 seconds.
To find the answers to these questions, we can use the equations of projectile motion.
a) To find the horizontal range, we need to determine the time of flight of the shell. The time of flight can be calculated using the equation:
Time of flight (T) = 2 * (initial vertical velocity) / (acceleration due to gravity)
However, we know the initial vertical velocity is given as 1.60 × 10^3 m/s and the acceleration due to gravity is 9.81 m/s^2.
Therefore:
T = 2 * (1.60 × 10^3) / (9.81)
Simplifying this equation, we get:
T = 324.77 s (approximately)
Now, we can calculate the horizontal range using the equation:
Horizontal range = (initial horizontal velocity) * (time of flight)
The initial horizontal velocity is given as: (initial velocity) * cos(angle)
Therefore:
Horizontal range = (1.60 × 10^3) * cos(43.0°) * (324.77)
Simplifying this equation, we get:
Horizontal range ≈ 2.41 × 10^5 m
So, the horizontal range of the shell is approximately 2.41 × 10^5 meters.
b) The time of flight of the shell is already calculated as 324.77 seconds (from part a). Therefore, we know that the shell is in motion for 324.77 seconds.
horizontal problem:
u = 1600 cos 43 the whole time
range = u t
where t is the time in the air
vertical problem to get t:
initial vertical speed = Vi = 1600 sin 43
v = Vi - g * time
at top, v = 0 and time = t/2
so
0 = 1600sin 43 - 9.81 (t/2)
so
t = (3200 sin 43 /9.81)
now go back and use that t in range = u t