Question:
IN THE TOPIC : PROJECTILE MOTION
YOU HAVE GIVEN A DERIVATION OF-:- MAXIMUM HEIGHT OF A PROJECTILE and it is ok but i had proved it by my method and i DO'NOT arrived at SAME RESULT but it is SIMILAR.
TELL ME WHERE I,am wrong.
method:-:
by formula {distance=speed x time)
here, speed=U SIN(THETA)
time=U SIN(THETA)/g
so,
d=u sin(theta) X u sin(theta)/g
=u(square) sin(square)(theta)/g
proved
your answer is 1/2 x u(square) sin(square)(theta)/g
please tell my mistake
Answer:
The maximum height, ymax, can be found from the equation:
vy 2 = voy2 + 2 ay (y - yo)
yo = 0, and, when the projectile is at the maximum height, vy = 0.
Solving the equation for ymax gives:
ymax = - voy2 /(2 ay)
Plugging in voy = vo sin(q) and ay = -g, gives:
ymax = vo2sin2(q) /(2 g)
where g = 9.8 m/s2
Note that the maximum height is determined solely by the initial velocity in the y direction and the acceleration due to gravity. Its not affected by whats happening in the x direction.
The maximum height, ymax, can be found from the equation:
vy 2 = voy2 + 2 ay (y - yo)
yo = 0, and, when the projectile is at the maximum height, vy = 0.
Solving the equation for ymax gives:
ymax = - voy2 /(2 ay)
Plugging in voy = vo sin(q) and ay = -g, gives:
ymax = vo2sin2(q) /(2 g)
where g = 9.8 m/s2
Note that the maximum height is determined solely by the initial velocity in the y direction and the acceleration due to gravity. Its not affected by whats happening in the x direction.
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