A projectile is fired with an initial speed of 61.3m/s at an angle of 44.2∘ above the horizontal on a long flat firing range.
(a) Determine the maximum height reached by the projectile.
(b) Determine the total time in the air.
(c) Determine the total horizontal distance covered (that is, the range).
(d) Determine the speed of the projectile 1.15s after firing.
Help! I have no idea where to begin or how to solve. Please explain!
To solve this problem, we can use the equations of projectile motion.
(a) To determine the maximum height reached by the projectile, we can use the fact that at the highest point of the projectile's trajectory, the vertical component of the velocity is zero.
We can find the vertical component of the initial velocity (Vi) using the trigonometric functions.
Vi = V * sin(θ)
where V is the initial speed of the projectile and θ is the angle above the horizontal.
Vi = 61.3 m/s * sin(44.2∘)
Vi ≈ 61.3 m/s * 0.6935
Vi ≈ 42.5 m/s
Next, we can use the equation for vertical motion:
Vy^2 = Vi^2 - 2 * g * Δy
where Vy is the vertical component of the velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and Δy is the change in height (which is what we're trying to find).
Since the projectile is at its highest point, Vy = 0. Therefore, the equation becomes:
0 = Vi^2 - 2 * g * Δy
Solving for Δy:
Δy = Vi^2 / (2 * g)
Δy = (42.5 m/s)^2 / (2 * 9.8 m/s^2)
Δy ≈ 925.54 m
So, the maximum height reached by the projectile is approximately 925.54 meters.
(b) To determine the total time in the air, we can use the equation for the vertical component of motion:
Vy = Vi - g * t
Since the projectile reaches its maximum height when Vy = 0, we can find the time it takes to reach the maximum height. Solving for t:
0 = Vi - g * t
t = Vi / g
t = 42.5 m/s / 9.8 m/s^2
t ≈ 4.34 s
However, this is only the time it takes for the projectile to reach its maximum height. The total time in the air is twice this time, as the projectile will take the same amount of time to come back down.
Total time in the air = 2 * t
Total time in the air ≈ 2 * 4.34 s
Total time in the air ≈ 8.68 s
So, the total time in the air is approximately 8.68 seconds.
(c) To determine the total horizontal distance covered (range), we can use the equation for horizontal motion:
Range = Vx * t
where Vx is the horizontal component of the velocity, which is given by:
Vx = V * cos(θ)
Range = V * cos(θ) * t
Range = 61.3 m/s * cos(44.2∘) * 8.68 s
Range ≈ 61.3 m/s * 0.7639 * 8.68 s
Range ≈ 420.55 m
So, the total horizontal distance covered (range) is approximately 420.55 meters.
(d) To determine the speed of the projectile 1.15 seconds after firing, we can use the equation for the horizontal component of motion:
Vx = V * cos(θ)
Vx = 61.3 m/s * cos(44.2∘)
Vx ≈ 61.3 m/s * 0.7639
Vx ≈ 46.88 m/s
Therefore, the speed of the projectile 1.15 seconds after firing is approximately 46.88 meters per second.
Sure! To solve this problem, we can use the equations of projectile motion. Let's break down each part of the question and find the solution step by step.
(a) To determine the maximum height reached by the projectile, we need to find the vertical component of its motion. We can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
In this case, the initial velocity (u) is 61.3 m/s, and the angle above the horizontal is 44.2 degrees. We can decompose the initial velocity into its vertical and horizontal components:
Vertical component: u_y = u * sin(theta)
Horizontal component: u_x = u * cos(theta)
Using g = 9.8 m/s^2 as the acceleration due to gravity, we can find the time it takes for the projectile to reach maximum height:
v_y = u_y - gt
0 = u_y - gt
Rearranging this equation gives us:
t = u_y / g
Now we can use this time to find the maximum height (h_max) by using the equation:
h_max = u_y * t - 0.5 * g * t^2
Substituting the values, we get:
h_max = (u * sin(theta)) * (u * sin(theta)) / (2 * g)
Substituting the given values, we have:
h_max = (61.3 * sin(44.2))^2 / (2 * 9.8)
Calculating that gives us:
h_max ≈ 40.876 meters
Therefore, the maximum height reached by the projectile is approximately 40.876 meters.
(b) To determine the total time in the air, we need to find the time it takes for the projectile to reach the ground after it is fired. We can use the equation of motion:
s = u * t + 0.5 * a * t^2
In this case, we are interested in the total time, so we can use the vertical motion equation:
s_y = u_y * t - 0.5 * g * t^2
Since the vertical displacement is zero at the end (when the projectile reaches the ground), we can solve for t:
0 = u_y * t - 0.5 * g * t^2
Simplifying the equation gives us:
0.5 * g * t^2 = u_y * t
t * (0.5 * g * t - u_y) = 0
This equation will have two solutions: t = 0 (at the start) and t = (2 * u_y) / g (at the end). We are interested in the latter solution:
t = (2 * u * sin(theta)) / g
Substituting the given values, we have:
t = (2 * 61.3 * sin(44.2)) / 9.8
Calculating that gives us:
t ≈ 8.17 seconds
Therefore, the total time in the air is approximately 8.17 seconds.
(c) To determine the total horizontal distance covered (the range), we can use the horizontal component of the motion. The range (R) is given by the equation:
R = u_x * t
Substituting the given values, we have:
R = (61.3 * cos(44.2)) * 8.17
Calculating that gives us:
R ≈ 362.58 meters
Therefore, the total horizontal distance covered (range) is approximately 362.58 meters.
(d) To determine the speed of the projectile 1.15 seconds after firing, we can use the horizontal and vertical components of the motion at that time.
Horizontal component: v_x = u_x
Vertical component: v_y = u_y - g * t
Substituting the given values, we have:
v_x = 61.3 * cos(44.2)
v_y = 61.3 * sin(44.2) - 9.8 * 1.15
Calculating that gives us:
v_x ≈ 43.8 m/s
v_y ≈ 25.9 m/s
To find the speed (v), we can use the Pythagorean theorem:
v = sqrt(v_x^2 + v_y^2)
Substituting the values, we have:
v = sqrt((43.8)^2 + (25.9)^2)
Calculating that gives us:
v ≈ 51.03 m/s
Therefore, the speed of the projectile 1.15 seconds after firing is approximately 51.03 m/s.
I hope this explanation helps you understand how to solve these projectile motion problems!