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A stone is projected from a horizontal ground. It attains maximum height 'H' & immediately strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assuming the collision to be elastic, the height of the point on the wall from the ground where ball will strike is- (A)H/2 (B)H/4 (C)3H/4 (D) None of these?
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A stone is projected from a horizontal ground. It attains maximum heig...
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A stone is projected from a horizontal ground. It attains maximum heig...
To find the maximum height H attained by the stone, we need to consider the projectile motion of the stone.

In projectile motion, the vertical motion can be analyzed separately from the horizontal motion.

The vertical motion of the stone can be described by the equation:

y = yo + Vyo*t - (1/2)gt^2

Where:
- y is the vertical position of the stone at time t
- yo is the initial vertical position of the stone
- Vyo is the initial vertical velocity of the stone
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time

At the maximum height H, the vertical velocity Vyo becomes zero. Therefore, the equation becomes:

0 = Vyo - gt

Solving for t, we get:

t = Vyo/g

Substituting this value of t into the equation for y, we get:

H = yo + Vyo*(Vyo/g) - (1/2)g*(Vyo/g)^2

Simplifying further:

H = yo + Vyo^2/(2g)

Thus, the maximum height H attained by the stone is given by the equation H = yo + Vyo^2/(2g), where yo is the initial vertical position of the stone and Vyo is the initial vertical velocity of the stone.
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A stone is projected from a horizontal ground. It attains maximum height 'H' & immediately strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assuming the collision to be elastic, the height of the point on the wall from the ground where ball will strike is- (A)H/2 (B)H/4 (C)3H/4 (D) None of these?
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