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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 - NEET (UG) - Physics

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Question

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?

Solution 1

Speed of the ball, u = 40 m/s

Maximum height, h = 25 m

In projectile motion, the maximum height reached by a body projected at an angle θ, is given by the relation:

h = `(u^2sin^2theta)/(2g)`

25 = `((40)^2sin^2theta)/(2xx9.8)`

sinθ = 0.30625

sin θ = 0.5534

θ = sin–1(0.5534) = 33.60°

Horizontal range, R  = `(u^2sin 2theta)/g`

= `((40)^2xxsin2xx33.60) /9.8`

=`(1600xxsin 67.2)/9.8`

=`(1600xx0.922)/9.8 = 150.53 m`

Solution 2

Maximum height h_"max" = 25 m; Horizonal range, R = ?

Velocity of projection, v = `40 ms^(-1)`

We know that `h_"max" = `(v^2sin^2 theta)/2g`

or `sin^2 theta = (25xx2xx9.8)/40xx40 = 0.30625 or `sin theta = 0.5534`

`theta = sin^(-1)(0.5534) = 33.6^@`

Again `R = (v^2sin 2theta)/g = (40xx40sin 67.2^@)/9.8 `

or `R = 1600/9.8 xx 0.9219 m = 150.5 m`

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Physics Textbook for Class 11 (2018 to Current)
Chapter 4: Motion in a Plane
Q: 15 | Page no. 87
Solution 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 Concept: Projectile Motion.
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