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

1

Answers

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

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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
========================
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
============================
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
===============================
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

  check your answer mentioned as 80 meters... is it correct ??

1 5 1
click on thanks button above please
Answer is 80 m
i am not sure..of that..
Okay sir