A projectile fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at an angle of elevation of 30 deg with the horizontal.
i. The greatest elevation above the ground reached by the projectile.
ii. Horizontal distance from the gun to the point, where the projectile strikes the ground.
vertical problem
Hi = 150 m
Vi = 180 sin 30 = 90 m/s
v = Vi - g t = 90 - 9.81 t
v = 0 at top so t = 9.17 seconds to top
h = Hi + Vi t - (g/2) t^2
h at top = 150 + 90(9.17) - 4.9(9.17^2)
= 150 + 825 - 412 = 563 meters above ground at top after 9.17 s
now how long before ground?
falls from 563 meters high
v = - g T
0 = 563 - 4.9 T^2
T = 10.7 seconds falling
total time = t + t = 9.17 + 10.7 = 19.9 seconds
horizontal speed = 180 cos 30 = 156 m/s
so 156 m/s for 19.9 seconds ---> 3102 meters horizontal to target
Given: 180m/s - initial velocity, 150m- height of cliff, 30° angle
Unknown: H- greatest elevation, R- horizontal distance until it strikes the ground
H=150m + (Vosin30°)² /2g
H=150 + (180m/s sin30°)² /2(9.81m/s²)
=562.844meters
R=(Vo² sin2(30°))/g
R=((180m/s)²sin(60°))/(9.71m/s²)
R=2860.267 meters
KEEP UP
i. Well, the greatest elevation above the ground reached by the projectile is the highest point it reaches during its flight. And if I had a dollar for every time I reached new heights, I'd have at least three dollars! Anyway, to find this point we can use the projectile motion equations. We know the initial velocity (180 m/s) and the launch angle (30 degrees), so we can split the velocity into horizontal and vertical components. The vertical component of velocity is 180 m/s * sin(30), which is 90 m/s. Now, we can use the kinematic equation for vertical motion:
v^2 = u^2 + 2as
Where v is the final vertical velocity (0 m/s at the top), u is the initial vertical velocity (90 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (which we want to find).
Substituting the known values, we get:
0^2 = (90)^2 + 2*(-9.8)*s
Simplifying further, we find:
-9.8s = -4050
Solving for s, we find that the displacement is approximately 414.3 meters.
So, the greatest elevation above the ground reached by the projectile is 414.3 meters!
ii. Now let's find the horizontal distance from the gun to the point where the projectile strikes the ground. This can be found using the horizontal component of the initial velocity, which is 180 m/s * cos(30), which is approximately 155.9 m/s.
Now, we can use the kinematic equation for horizontal motion:
s = ut
Where s is the horizontal displacement, u is the initial horizontal velocity, and t is the time of flight.
To find the time of flight, we can use the equation:
s = ut + (1/2)at^2
Since the horizontal acceleration is 0 (no horizontal forces acting on the projectile), the equation simplifies to:
s = ut
Substituting the known values, we have:
s = (155.9)(t)
To find t, we can use the vertical motion equation:
s = ut + (1/2)at^2
But this time, we're looking for the time it takes for the projectile to hit the ground, so the final vertical displacement is 0.
Thus, we have:
0 = (90)t + (1/2)(-9.8)t^2
Simplifying further, we find:
-4.9t^2 + 90t = 0
Factoring out t, we get:
t(-4.9t + 90) = 0
So, t = 0 or t = 90/4.9
Since we're looking for the positive time (ignoring the t = 0 solution), we find that t ≈ 18.4 seconds.
Now we can substitute this value of t back into the equation for horizontal motion:
s = (155.9)(t)
Substituting t = 18.4, we find:
s ≈ 2867 meters
So, the horizontal distance from the gun to the point where the projectile strikes the ground is approximately 2867 meters!
To find the answers to these questions, we can use the equations of motion for projectile motion.
i. The greatest elevation above the ground reached by the projectile:
In projectile motion, the vertical motion is influenced by gravity, while the horizontal motion remains constant. The time of flight for the projectile can be found using the equation:
time = (2 * initial vertical velocity * sin(angle)) / acceleration due to gravity
Here, the initial vertical velocity is given by:
initial vertical velocity = initial velocity * sin(angle)
The maximum height reached by the projectile can be calculated using the equation:
maximum height = (initial vertical velocity^2) / (2 * acceleration due to gravity)
Plugging in the given values, we can calculate the maximum height.
ii. The horizontal distance from the gun to the point where the projectile strikes the ground:
To calculate the horizontal distance, we need to find the time of flight of the projectile. The equation for time of flight is:
time = (2 * initial vertical velocity * sin(angle)) / acceleration due to gravity
Once we have the time of flight, we can calculate the horizontal distance using the equation:
horizontal distance = initial velocity * cos(angle) * time
Now, let's plug in the given values and find the answers:
Given:
Height of the cliff (h) = 150 m
Initial velocity (v) = 180 m/s
Angle of elevation (θ) = 30 degrees
Acceleration due to gravity (g) = 9.8 m/s²
i. The greatest elevation above the ground reached by the projectile:
Calculate the initial vertical velocity:
initial vertical velocity = v * sin(θ) = 180 m/s * sin(30) = 90 m/s
Calculate the maximum height:
maximum height = (initial vertical velocity^2) / (2 * g) = (90 m/s)^2 / (2 * 9.8 m/s²) = 411.24 m
Therefore, the greatest elevation above the ground reached by the projectile is 411.24 m.
ii. The horizontal distance from the gun to the point where the projectile strikes the ground:
Calculate the time of flight:
time = (2 * initial vertical velocity * sin(angle)) / g = (2 * 90 m/s * sin(30)) / 9.8 m/s² = 11.53 s
Calculate the horizontal distance:
horizontal distance = initial velocity * cos(angle) * time = 180 m/s * cos(30) * 11.53 s = 1982.86 m
Therefore, the horizontal distance from the gun to the point where the projectile strikes the ground is 1982.86 m.