A ball is projected vertically upwardsits speed at half of maximum height is 30m/s maximum height attained by the ball is
Answers
Answer:
Max Height = 90 m
Vh = V/√2
Explanation:
A ball is projected vertically upwardsits speed at half of maximum height is 30m/s maximum height attained by the ball is
Let Say Ball is thrown With Velocity V
Speed at Max height = 0
V² - U² = 2aS
=> 0² -V² = 2(-g)H
=> V² = 2gH
=> H = V²/2g
Max Height = H
Velocity at H/2 = Vh
Vh² - U² = 2a(H/2)
=> Vh² - V² = -2gH/2
=> Vh² - V² = -V²/2
=> Vh² = V²/2
=> Vh = V/√2
Speed at half of max height would be 1/√2 times of Projected Speed
Vh = 30 m
V = 30√2
H = V²/2g = (30√2)²/2*10 = 90 m
Max Height = 90 m
Let maximum height attained by ball is H.
the ball is projected vertically upwards and its speed at half of maximum height (i.e., H/2) is 30m/s.
let velocity of ball at the height H/2 from the ground = initial velocity of ball , u = 30m/s
we know, velocity at maximum height becomes zero.
so, final velocity of ball , v = velocity at maximum height = 0
now using formula, v² = u² + 2as
here , v = 0, u = 30 , a = -g = -10m/s² and s = H/2
so, 0² = 30² + 2(-10) × H/2
or, H = 90 m
hence, maximum height attained by ball is 90m