A man stands on the roof of a 10.0 -tall building and throws a rock with a velocity of magnitude 30.0 at an angle of 38.0 above the horizontal. You can ignore air resistance.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
This problem was posted before, perhaps by you, since "10.0-" is still written instead of 10.0 meters. You need to say what the dimensions are.
As I said in my previous answer, the kinetic energy increase by M g H, where H = 10 meters. Since kinetic energy is (1/2) m V^2, you can divide out the Mass and say that V^2 increases by gH. g is the acceleration of gravity.
I meant to write (as I did when you asked the question before):
V^2 increases by 2gH.
g is the acceleration of gravity.
Therefore
Vfinal^2 = 30^2 + 2*9.8*10 = 1096 m^2/s^2
Vfinal = 33.1 m/s
To calculate the horizontal distance from the base of the building to the point where the rock strikes the ground, we can use the kinematic equations of projectile motion.
First, let's break down the initial velocity of the rock into its horizontal and vertical components. The horizontal component (Vx) can be calculated as:
Vx = V * cos(theta)
= 30.0 * cos(38.0)
= 23.12 m/s
The vertical component (Vy) can be calculated as:
Vy = V * sin(theta)
= 30.0 * sin(38.0)
= 18.27 m/s
Since the rock is thrown vertically above the horizontal, the initial vertical velocity is positive. However, due to gravity, the vertical velocity will change over time, and eventually become negative when the rock starts to fall back down.
Next, we can calculate the time of flight (t) for the rock to reach the ground. Since we know the initial vertical velocity and the height of the building, we can use the equation:
Vy = Voy - g * t
where Voy is the initial vertical velocity (18.27 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
Rearranging the equation and solving for t, we get:
t = (Vy - Voy) / g
= (0 - 18.27) / (-9.8)
= 1.86 s
Now that we have the time of flight, we can calculate the horizontal distance (d) using the equation:
d = Vx * t
= 23.12 * 1.86
= 42.97 m
Therefore, the horizontal distance from the base of the building to the point where the rock strikes the ground is approximately 42.97 meters.