A body is projected horizontally with a velocity of 80m/s from the top of a tower 160m above the ground (given that g=10m/s2).Find (1)time of flight. (2)Range
time of flight:
h=1/2 g t^2
t=sqrt (2h/g)
horizontal range=80*t
To find the time of flight and range of a body projected horizontally, we can use the equations of motion and the kinematic equations. Let's go step by step to find the answers:
1. Time of Flight:
The time of flight is the total time taken by the body to reach the ground. Since the body is projected horizontally, there is no initial vertical velocity, and the only force acting on it is gravity. We can use the equation for vertical motion:
h = (1/2) * g * t^2
where:
h = vertical displacement (160m)
g = acceleration due to gravity (10m/s^2)
t = time of flight (what we need to find)
Rearranging the equation, we get:
t^2 = (2h) / g
t^2 = (2 * 160) / 10
t^2 = 32
t = square root of 32
t ≈ 5.657 seconds
Therefore, the time of flight is approximately 5.657 seconds.
2. Range:
The range is the horizontal distance covered by the body during its time of flight. Since the body is projected horizontally, its horizontal velocity remains constant throughout the motion. We can calculate the range using the formula:
Range = horizontal velocity * time of flight
Given:
Horizontal velocity = 80 m/s
Time of flight = 5.657 seconds (from previous calculation)
Range = 80 * 5.657
Range ≈ 452.56 meters
Therefore, the range of the body is approximately 452.56 meters.
Note: This calculation assumes that air resistance is negligible.