Particle is dropped from the height of 20 m from ground. A constant force acts on the particle in horizontal direction due to which acceleration of the particle become 6m/s. Find the horizontal displacement of the particle till it reaches ground
Answer is 12 m
Please solve this numerical
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the equations that are relevant are : (in horizontal and vertical directions)
v = u + a t
v² = u² + 2 a s
s = (v +u)/2 * t
s = u t + 1/2 a t²
Horizontal acceleration does not affect the flight in the vertical direction.
vertical flight:
height h = u t + 1/2 g t²
20 = 0 + 1/2 g * t² => time of flight = t² = 20 * 2 / 10
t = 2 sec.
In 10 seconds, the particle will travel horizontally: The displacement is :
s = u t + 1/2 a t² = 0 + 1/2 * 6 * 2² = 12 meters
v = u + a t
v² = u² + 2 a s
s = (v +u)/2 * t
s = u t + 1/2 a t²
Horizontal acceleration does not affect the flight in the vertical direction.
vertical flight:
height h = u t + 1/2 g t²
20 = 0 + 1/2 g * t² => time of flight = t² = 20 * 2 / 10
t = 2 sec.
In 10 seconds, the particle will travel horizontally: The displacement is :
s = u t + 1/2 a t² = 0 + 1/2 * 6 * 2² = 12 meters
kvnmurty:
click on thanks button above please
I forget to put initial velocity of particle zero
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