A ball is thrown at an angle 30 from the horizontal with a velocity 20...
Problem Statement
A ball is thrown at an angle 30 from the horizontal with a velocity 20m/s from the top of a building of height 40m . Find its time of flight and horizontal range?
Solution
Step 1: Break down the given information
- Angle of projection: 30 degrees
- Initial velocity: 20 m/s
- Height of building: 40 m
Step 2: Calculate time of flight
Time of flight can be calculated using the formula:
Time of flight = (2 * u * sin(theta)) / g
- u = initial velocity = 20 m/s
- theta = angle of projection = 30 degrees
- g = acceleration due to gravity = 9.8 m/s^2
Therefore, substituting the values in the formula:
Time of flight = (2 * 20 * sin(30)) / 9.8 = 4.08 seconds
Step 3: Calculate horizontal range
Horizontal range can be calculated using the formula:
Horizontal range = u^2 * sin(2 * theta) / g
- u = initial velocity = 20 m/s
- theta = angle of projection = 30 degrees
- g = acceleration due to gravity = 9.8 m/s^2
Therefore, substituting the values in the formula:
Horizontal range = 20^2 * sin(2 * 30) / 9.8 = 58.24 meters
Step 4: Interpretation of the results
The ball would take 4.08 seconds to reach the ground and would travel a horizontal distance of 58.24 meters. This means that the ball would land approximately 58.24 meters away from the building.