Consider a rock that is thrown off a bridge of height 27 m at an angle θ = 24° with respect to the horizontal as shown in the figure above. If the initial speed the rock is thrown is 12 m/s, find the following quantities:

(a) The time it takes the rock to reach its maximum height.
s

(b) The maximum height reached by the rock.
m

(c) The time at which the rock lands.
s

(d) The place where the rock lands.
m

(e) The velocity of the rock (magnitude and direction) just before it lands.
Magnitude Direction
m/s °

Frank, Jim, Tommy: It appears you are answer grazing under assumed names. Stop it, please. Show some work or thinking.

To find the answers to the given questions, we can use the kinematic equations of motion. These equations relate the initial velocity, final velocity, acceleration, time, and displacement of an object. In this case, we'll apply these equations to the vertical motion of the rock.

(a) To find the time it takes the rock to reach its maximum height, we'll focus on the vertical motion. We know that the initial vertical velocity is given by V_initial = V0 * sin(θ), where V0 is the initial speed and θ is the angle. The equation for vertical displacement can be given as s = V_initial * t + 1/2 * a * t^2, where s is the displacement, a is the acceleration (which is -9.8 m/s^2 due to gravity), and t is the time. At the maximum height, the vertical velocity becomes zero, so V_final = V_initial + a * t. Solving for t, we can find the time it takes to reach the maximum height.

(b) The maximum height reached by the rock occurs when the vertical velocity is zero, so the equation becomes V_final = –V0 * sin(θ) + a * t. Using this equation, we can find the displacement at the maximum height.

(c) To find the time at which the rock lands, we need to consider the entire vertical motion. The equation for vertical displacement can be used again, where we know the displacement is equal to the negative of the initial height (s = -27 m). Solving for t, we'll be able to find the time at which the rock lands.

(d) Knowing the time at which the rock lands, we can use the horizontal motion to find the horizontal displacement. The horizontal velocity is given by V_horizontal = V0 * cos(θ), and the horizontal displacement is given by s_horizontal = V_horizontal * t.

(e) Finally, to find the velocity of the rock just before it lands, we can use the vertical and horizontal velocities at that time. The magnitude of the velocity is given by V = sqrt(V_horizontal^2 + V_vertical^2), and the direction is given by atan(V_vertical / V_horizontal).

Now, let's plug in the values and calculations into these equations to find the answers:

(a) The time it takes the rock to reach its maximum height:
V_initial = 12 m/s * sin(24°)
t = -V_initial / a

(b) The maximum height reached by the rock:
V_final = -V_initial + a * t
s = V_initial * t + 1/2 * a * t^2

(c) The time at which the rock lands:
s = -27 m
t = sqrt((2 * s) / a)

(d) The place where the rock lands:
V_horizontal = 12 m/s * cos(24°)
s_horizontal = V_horizontal * t

(e) The velocity of the rock (magnitude and direction) just before it lands:
V_velocity = -V0 * sin(θ) + a * t
V = sqrt(V_horizontal^2 + V_velocity^2)
Direction = atan(V_velocity / V_horizontal)

Now, by substituting the values and evaluating the calculations, we can find the answers.