The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s ¯¹ can go without hitting the ceiling of the hall ?
Given: The velocity of the ball is 40 ms -1 and the height of ceiling is 25 m .
The maximum height reached by a body in projectile motion is given as,
h = u 2 sin 2 θ 2 g
Where, initial velocity of ball is u and angle of projection of the ball is θ .
By substituting the given values in the above expression, we get
25 = ( 40 ) 2 sin 2 θ 2 × 9.8 sin 2 θ = 25 × 2 × 9.8 ( 40 ) 2 = 0.30625 sin θ = 0.30625
Again solving for θ , we get
θ = sin − 1 ( 0.5534 ) = 33.60 °
The horizontal range is given as,
R = u 2 sin 2 θ g
By substituting the given values in the above formula, we get
R = ( 40 ) 2 sin ( 2 × 33.6 ) 9.8 = 150.5 m
Thus, the ball covers a horizontal distance of 150.5 m before hitting the ceiling.
Given: The velocity of the ball is
The maximum height reached by a body in projectile motion is given as,
Where, initial velocity of ball is
By substituting the given values in the above expression, we get
Again solving for
The horizontal range is given as,
By substituting the given values in the above formula, we get
Thus, the ball covers a horizontal distance of
