A 0.54 kg rock is projected from the edge of the top of a building with an initial velocity of 13.8 m/s at an angle 59◦ above the horizontal.Due to gravity, the rock strikes the ground at a horizontal distance of 20.3 m from the base of the building.How tall is the building? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is
9.8 m/s^2. Answer in units of m
To find the height of the building, we need to break down the motion of the rock into its horizontal and vertical components.
First, let's find the time it takes for the rock to reach the ground. We know that the horizontal distance traveled by the rock is 20.3 m, and the initial horizontal velocity is 13.8 m/s. The horizontal component of motion is not affected by gravity, so we can use the formula:
distance = velocity × time
Solving for time:
20.3 m = 13.8 m/s × time
time = 20.3 m / 13.8 m/s
Now, we can find the vertical distance the rock travels using the time we just calculated. In the vertical direction, the initial velocity is 13.8 m/s at an angle of 59 degrees above the horizontal. We can break this velocity into its vertical and horizontal components using trigonometry.
The initial vertical velocity is given by:
vertical velocity = initial velocity * sine(angle)
vertical velocity = 13.8 m/s * sin(59 degrees)
Now we can use the equation of motion for vertical motion:
vertical distance = (initial vertical velocity × time) + (0.5 × acceleration × time^2)
The initial vertical velocity is the value we just calculated, and the acceleration is due to gravity (9.8 m/s^2). The time is the same value we found earlier.
Finally, the height of the building is equal to the vertical distance traveled by the rock.
the vertical component of the initial velocity is 13.8 sin59° = 11.83 m/s
the horizontal component is
So, the height y is 7.11 m/s
since the rock hit 20.3 m away, it fell for 20.3/7.11 = 2.855 sec
h + 11.83*2.855 - 4.9*2.855^2 = 0
h = 6.165 m