A projectile is fired with an initial speed of 61.3m/s at an angle of 31.5° 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 1.44s after firing.
e)Determine the direction of the projectile 1.44s after firing.
a
vertical problem
vi = 61.3 sin 31.5 = 32 m/s up
when at top (v = 0) ?
o = 32 - 9.81 t
t = 3.26 seconds to max ht
how high?
h = Vi t - (1/2)(9.81)t^2
h = 32 (3.26) - 4.9(10.7)
= 51.9 m high
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easier way by the way:
average speed up = 32/2 = 16m/s for 3.26 seconds --> 52 m
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b
time in air = twice the time up = 3.26*2 = 6.52 seconds
c
horizontal problem
constant horizontal speed = 61.3 cos 31.5 = 52.3 m/s
time is still 6.52 s
range = 52.3 * 6.52 = 341 meters
d
vertical problem
v = 32 - 9.81 (1.44)
= 17.9 m/s
horizontal speed is always 52.3
s = sqrt(17.9^2+2.3)^2
e
tan angle up from hor = 17.9/52.3
= 18.9 degrees up from horizontal
To solve this problem, we need to use the equations of motion for projectile motion. Let's break down each part of the problem step by step:
a) To determine the maximum height reached by the projectile, we can use the equation for vertical motion:
vf = vi + gt
where vf is the final vertical velocity, vi is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
At the maximum height, the final velocity will be zero, so we can set vf = 0. Since the initial velocity is given as 61.3 m/s, the vertical acceleration due to gravity is 9.8 m/s^2, and the angle of projection is 31.5 degrees (above the horizontal), we can break down the initial velocity into its vertical and horizontal components:
vi_y = vi * sin(theta)
= 61.3 m/s * sin(31.5°)
Using this vertical component, we can find the time it takes for the projectile to reach its maximum height:
0 = vi_y + gt_max
t_max = -vi_y / g
Plugging in the values, we get:
t_max = - (61.3 m/s * sin(31.5°)) / (9.8 m/s^2)
To find the maximum height, we can use the equation:
h_max = vi_y * t_max + 0.5 * g * t_max^2
Substituting the values, we get:
h_max = (61.3 m/s * sin(31.5°)) * t_max + 0.5 * 9.8 m/s^2 * t_max^2
b) The total time in the air can be found by using the equation:
t_total = 2 * t_max
c) The total horizontal distance covered (range) can be found using the equation:
R = vi_x * t_total
where vi_x is the initial horizontal velocity.
vi_x = vi * cos(theta)
= 61.3 m/s * cos(31.5°)
d) To find the speed of the projectile 1.44 s after firing, we can use the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
where vx is the horizontal velocity and vy is the vertical velocity at 1.44 s.
vx = vi_x
vy = vi_y + g * t
= vi * sin(theta) + g * t
e) The direction of the projectile 1.44 s after firing can be determined by finding the angle θ with the horizontal:
θ = atan(vy / vx)
To solve these problems, we can use the principles of projectile motion and apply the equations of motion in both the horizontal and vertical directions. Here's how we can approach each part of the question:
a) To find the maximum height reached by the projectile, we need to determine its vertical displacement. We can use the equation:
Δy = (v₀y)² / (2g)
where Δy represents the vertical displacement, v₀y is the initial vertical component of the velocity, and g is the acceleration due to gravity. The initial vertical velocity can be found using the formula:
v₀y = v₀ * sin(θ)
where v₀ is the initial speed of the projectile and θ is the launch angle (31.5° in this case).
By substituting the given values into the equations, we can calculate the maximum height.
b) To find the total time in the air, we can use the equation:
t = 2 * v₀y / g
In this case, we already know the initial vertical velocity (v₀y) and the acceleration due to gravity (g). By plugging in these values, we can determine the total time the projectile spends in the air.
c) To find the total horizontal distance covered, we can use the equation:
R = v₀x * t
where R is the total range, v₀x is the initial horizontal component of the velocity, and t is the total time in the air. The initial horizontal velocity can be found using the formula:
v₀x = v₀ * cos(θ)
By substituting the given values into the equations, we can calculate the total horizontal distance.
d) To find the speed of the projectile 1.44 seconds after firing, we need to decompose the initial velocity into its horizontal and vertical components. We can then use the equations of motion to find the velocity components at any given time. The vertical component of the velocity remains constant due to the acceleration of gravity, while the horizontal component remains constant since there are no horizontal forces acting on the projectile.
By using the time elapsed after firing (1.44 seconds) and the given launch angle, we can calculate the new vertical and horizontal velocities. The speed of the projectile can be found using the equation:
v = sqrt((v_x)² + (v_y)²)
e) To determine the direction of the projectile 1.44 seconds after firing, we can examine the signs of the horizontal and vertical components of the velocity. The sign of the horizontal velocity will indicate the direction of the projectile. If it is positive, the projectile is moving in the positive x-direction. If it is negative, the projectile is moving in the negative x-direction.