a body is projected horizontally with a velocity of 80m/s from the top of a tower 160m above the ground. find the time of flight,range,velocity at which the body strikes the ground
Vertical problem is falls 160 meters with no initial vertical speed, time same as a rock falling
so
a = -9.81 m/s^2
v = 0 - 9.81 t
0 = 160 + 0 t - (1/2)(9.81)t^2
so
t = sqrt (320/4.9) = 8.08 seconds falling
Now horizontal
no horizontal force so no change in horizontal speed
u = 80 m/s
d = 80 m/s * 8.08 s
about 646 meters on earth
the final velocity is the square root
... of the sum of the squares of the horizontal and vertical components
horizontal = 80 m/s
vertical = g * flight time ... or ā(2 * g * 160)
I dont no.
physics
To find the time of flight, range, and velocity at which the body strikes the ground, we need to use the equations of motion.
1. Time of Flight:
The time of flight is the total time taken by the body to reach the ground. We can calculate it using the equation:
š = ā(2ā/š)
Where:
š is the time of flight,
ā is the height of the tower (160m),
š is the acceleration due to gravity (approx. 9.8m/sĀ²).
Plugging in the values:
š = ā(2 * 160 / 9.8) ā 5.12 seconds
So, the time of flight is approximately 5.12 seconds.
2. Range:
The range is the horizontal distance covered by the body. Since the body is projected horizontally, the initial vertical velocity is zero. We can calculate the range using the equation:
š
= šš„ * š
Where:
š
is the range,
šš„ is the horizontal component of the velocity,
š is the time of flight.
The horizontal component of velocity (šš„) is 80 m/s since the body is projected horizontally.
š
= 80 * 5.12 ā 409.6 meters
So, the range is approximately 409.6 meters.
3. Velocity at which the body strikes the ground:
Since there is no vertical acceleration, the vertical component of velocity remains constant throughout the motion. The vertical component of velocity at the time of striking the ground is:
šš¦ = 0 + š * š
Where:
šš¦ is the vertical component of velocity,
š is the acceleration due to gravity (approx. 9.8m/sĀ²),
š is the time of flight.
Plugging in the values:
šš¦ = 9.8 * 5.12 ā 50.18 m/s
The horizontal component of velocity remains constant throughout the motion and is equal to the initial horizontal velocity (šš„ = 80 m/s).
So, the velocity at which the body strikes the ground is approximately 80 m/s horizontally and 50.18 m/s vertically.