A projectile is shot from the edge of a cliff 120 m above ground level with an initial speed of 60 m/s at an angle of 30.0o with the horizontal, as shown below. (a) How long does it take for the projectile to hit the ground? (b) How far does the projectile land from the base of the cliff? (c) What are the horizontal and the vertical components of its velocity when it hits the ground? (d) What are the magnitude of the velocity, and the angle made by the velocity vector with the horizontal. (e) What’s the maximum height above the cliff reached by the projectile. (f) How long does it take for the projectile to reach the maximum height?
Vertical problem:
Vi = 60 sin 30 = 30 m/s upwards
Hi = 120 meters high
a = g = -9.81 m/s^2
so
v = Vi - g t = 30 - 9.81 t
h = Hi + Vi t - (1/2) a t^2 = 120 + 30 t - 4.9 t^2
when does h = 0 (ground) ?
4.9 t^2 - 30 t - 120 = 0
solve quadratic
(a) t = +8.88 seconds or -2.76 seconds
use the +8.88. the negative answer was on the way up if we had shot it before t = 0
(b) u = horizontal speed = 60 cos 30 = 52m/s forever and ever while above ground
so horizontal distance = u t = 52 *8.88 = 461 meters
(c) v = Vi - g t = 30 - 9.81 * 8.88 = -57.1 m/s
and of course u = 52 m/s
(d) speed = sqrt (52^2+57.1^2)
tangent of angle below horizontal = 57.1 / 52
(e) and (f) back to part (a)
v = Vi - g t = 30 - 9.81 t
when is v = 0 (at the top)
t = 30/9.81 = 3.06 seconds to top
h above cliff = Vi t - 4.9 t^2 = 30(3.06) - 4.9(3.06)^2
= 91.8 - 45.9 = 45.9 meters above the cliff
To solve this problem, we can break it down into several steps:
Step 1: Solve for the time it takes for the projectile to hit the ground
Step 2: Calculate the horizontal distance traveled by the projectile
Step 3: Find the horizontal and vertical components of its velocity at impact
Step 4: Determine the magnitude and angle of the velocity vector at impact
Step 5: Calculate the maximum height reached by the projectile
Step 6: Find the time it takes for the projectile to reach the maximum height
Let's go through each step one by one:
Step 1: Solve for the time it takes for the projectile to hit the ground
We'll use the kinematic equation for vertical motion:
y = v₀y * t + (1/2) * a * t²
where:
y = -120 m (negative because it's below the starting point)
v₀y = v₀ * sin(θ) = 60 m/s * sin(30°)
a = -9.8 m/s² (acceleration due to gravity)
Substituting these values, we get:
-120 = 30 * t - 4.9 * t²
Rearranging the equation:
4.9 * t² - 30 * t - 120 = 0
Now we can solve this quadratic equation for t.
Step 2: Calculate the horizontal distance traveled by the projectile
To find the horizontal distance, we can use the kinematic equation for horizontal motion:
x = v₀x * t
where:
v₀x = v₀ * cos(θ) = 60 m/s * cos(30°)
t is the time calculated in Step 1.
Step 3: Find the horizontal and vertical components of its velocity at impact
The horizontal component of velocity remains constant throughout the projectile's motion, so it will be equal to v₀x.
The vertical component of velocity at impact can be found using the equation:
v_y = v₀y + a * t
where v₀y and t are the values calculated in Step 1.
Step 4: Determine the magnitude and angle of the velocity vector at impact
The magnitude of the velocity vector can be calculated using:
v = √(v_x² + v_y²)
where v_x and v_y are the horizontal and vertical components of velocity calculated in Step 3.
The angle made by the velocity vector with the horizontal can be found using:
θ_v = arctan(v_y / v_x)
Step 5: Calculate the maximum height reached by the projectile
The maximum height can be found using the vertical motion equation:
y_max = (v₀y²) / (2 * g)
where v₀y is the initial vertical speed and g is the acceleration due to gravity.
Step 6: Find the time it takes for the projectile to reach the maximum height
To find the time taken to reach the maximum height, we can use the equation:
v_y = v₀y + a * t_max
where v_y is the vertical component of velocity, v₀y is the initial vertical speed, a is the gravitational acceleration, and t_max is the time taken to reach the maximum height.
Now, let's calculate the values step by step.
To solve this problem, we can use the equations of motion for projectile motion. The key components we need to consider are the initial velocity, the angle of projection, the vertical and horizontal components of the projectile's motion, and the acceleration due to gravity.
(a) To determine how long it takes for the projectile to hit the ground, we can use the vertical motion equation:
h = v0y * t + (1/2) * g * t^2
where h is the initial height above the ground (120 m), v0y is the initial vertical component of velocity, t is the time taken, and g is the acceleration due to gravity (-9.8 m/s^2).
The initial vertical component of velocity can be calculated by multiplying the initial speed (60 m/s) by the sine of the angle of projection (30 degrees):
v0y = v0 * sin(theta)
Plugging these values into the equation, we can solve for t:
120 = (60 * sin(30)) * t + (1/2) * (-9.8) * t^2
Simplifying and solving this quadratic equation will give us the time it takes for the projectile to hit the ground.
(b) To determine how far the projectile lands from the base of the cliff (horizontal displacement), we can use the horizontal motion equation:
d = v0x * t
where d is the horizontal displacement, v0x is the initial horizontal component of velocity, and t is the time taken.
The initial horizontal component of velocity can be calculated by multiplying the initial speed (60 m/s) by the cosine of the angle of projection (30 degrees):
v0x = v0 * cos(theta)
Plugging in the values, we can solve for d.
(c) When the projectile hits the ground, the vertical velocity component is 0 because it is at its maximum height. The horizontal velocity component remains constant throughout the motion. Therefore, when it hits the ground, the horizontal component of its velocity will be the same as its initial horizontal component (v0x), and the vertical component of its velocity will be 0.
(d) To determine the magnitude of the velocity and the angle made by the velocity vector with the horizontal, we can use the equations:
magnitude of velocity (V) = sqrt(vx^2 + vy^2)
angle made with horizontal (θ) = atan(vy / vx)
where vx is the horizontal component of velocity and vy is the vertical component of velocity. We already know the values of vx (v0x) and vy (0) when it hits the ground, so we can calculate these.
(e) To find the maximum height above the cliff reached by the projectile, we can use the equation:
h_max = (v0y^2) / (2 * g)
Plugging in the values, we can solve for h_max.
(f) Lastly, to determine how long it takes for the projectile to reach the maximum height, we can use the equation:
vfy = v0y + g * t_max
where vfy is the final vertical component of velocity (0 when it reaches maximum height), v0y is the initial vertical component of velocity, g is the acceleration due to gravity, and t_max is the time taken to reach maximum height. We already know the values of vfy (0) and v0y when it hits the ground, so we can calculate t_max.
By using these equations and plugging in the given values, we can find the answers to the various parts of the question.