# A ball is rolled off the edge of a horizontal table at a speed of 4 m/s. It hits the ground after 0.4 s. Then find 1)horizontal distance covered by ball from table 2) height of table Q2 A bomber plane move horizontally with a speed 500 m/s and a bomb released from it, strike the ground in 10 sec. Angle of velocity vector with horizontal at which it strikes the ground Answer is tan-1(1/5) Q3 The speed of projectile at the highest point become 1/root 2times its initial speed The horizontal range of the projectile will be (if initial velocity is u) Answer is u square over get Q4 A cricketer can throw a ball to a maximum horizontal distance of 100 m.with same effort, he throws the ball vertical upward. The maximum height attained by ball is Answer is 80m

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by kaushikravikant

2015-09-11T22:04:19+05:30

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Q1)
let the height of the table be  h meters.  Initial vertical velocity of the ball is = u = 0.
the ball has a constant speed of 4 m/s in the horizontal direction.  It is accelerated by g in the vertically downward direction.

h = u t + 1/2 g t²              let g = 10 m/s²
= 0 + 1/2 * 10 * 0.4² = 0.8 meters

The horizontal distance covered by the ball from table in 0.4 sec
= 4 m/s * 0.4 sec = 1.6 meters
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Q2)
v = 500 m/s  horizontally
velocity in vertical direction when the ball hits the ground = w
w = u + g t = 0 + 10 m/s² * 10 s = 100 m/s
Let the angle the velocity (resultant) makes with horizontal be Ф.
tan Ф = vertical component / horizontal component
= 100 / 500 = 0.20
Ф = tan⁻¹ 0.20
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Q3)  let Ф be the angle of projection.    u = magnitude of velocity at the point of projection.
Given  u CosФ = velocity (magnitude) at the highest point = u /√2
CosФ = 1/√2  =>  Ф = π/4        => Sin Ф = 1/√2
The time taken by the projectile to reach the highest point = u SinФ/g
The horizontal range = R = 2 * (u cosФ) * (u SinФ / g) = u² / g
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Q4)
maximum Range = R = u² / g    as angle of projection = Ф = π/4 and sin 2 Ф = 1
So  u² / g = 100 m =>  u = 10√10 m/sec

If the ball is thrown vertically with the same energy and speed u, then
height reached h = (0² - u²) / (-2 g)  = 1000 /20 = 50 meters