A man stands on the roof of a building of height 13.8 and throws a rock with a velocity of magnitude 29.3 at an angle of 34.8 above the horizontal. You can ignore air resistance
OK, I will ignore it.
Now what is the question?
To determine the projectile motion of the rock, we can break down the initial velocity into its horizontal and vertical components.
The vertical component of the velocity can be given as:
V_y = V * sin(θ)
Where V is the magnitude of the velocity (29.3 m/s) and θ is the angle above the horizontal (34.8 degrees).
Substituting the given values:
V_y = 29.3 * sin(34.8)
Similarly, the horizontal component of the velocity can be given as:
V_x = V * cos(θ)
Where V is the magnitude of the velocity (29.3 m/s) and θ is the angle above the horizontal (34.8 degrees).
Substituting the given values:
V_x = 29.3 * cos(34.8)
The time of flight can be calculated as the time it takes for the rock to reach the ground from its initial height. The formula for the time of flight (t) is given by:
t = 2 * V_y / g
Where g is the acceleration due to gravity (9.8 m/s^2).
Substituting the values:
t = 2 * (29.3 * sin(34.8)) / 9.8
The horizontal distance traveled by the rock can be calculated using the formula:
Range = V_x * t
Substituting the previously calculated values for V_x and t:
Range = (29.3 * cos(34.8)) * t
To find the maximum height reached by the rock, we can use the formula for vertical displacement:
Vertical displacement = (V_y^2) / (2 * g)
Substituting the previously calculated value for V_y:
Vertical displacement = (29.3 * sin(34.8))^2 / (2 * 9.8)
By substituting the given values and evaluating the formulas, you can find the time of flight, horizontal distance, and maximum height reached by the rock.