# 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 can go without hitting the ceiling of the hall ?

the maximum possible height a projectile attains is given as

H = u2sin2θ / g

or the angle of projection will be given as

sinθ = [Hg / u2]1/2

here,

H = 25m

u = 40 m/s

g = 10 m/s2

so,

sinθ = [(25x10) / 402]1/2

or

sinθ = 0.559

thus,

θ = 34 degree

now,

the maximum horizontal range will be

R = u2sin2θ / g

= 402sin(2x34) / 10

thus, we get

R = 148.3 m

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• -31

Sorry i misunderstood the question...

4.15 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 ?
4.15
Solution:
Given:
Initial velocity of ball,u=40 ms-1 Maximum height ,H=25 m Acceleration due to gravity=9.8 ms-2
How,for a projectile maximum height acheived,H= u2 sin2θ 2g
⇒25=(40)2sin2θ 2×9.8
⇒sin2θ =0.3063
⇒sinθ= 0.3063=0.5534
⇒cosθ = 1−sin2θ = 1−0.3063=0.8329

• -32
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