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`        A train is moving on a track at 30 m/s. A ball is thrown from it perpendicular to the direction of motion at 30 m/s at 45° from horizontal. Find the distance of ball from the point of projection, when it strikes the ground `
one month ago

```							The initial velocity (in horizontal direction) will be having two components. One is in the direction of train and the other one is perpendicular to it. Therefore the resultant horizontal initial velocity of the ball is, Ux=(30 cos 45)2+302−−−−−−−−−−−−−−−√=450+900−−−−−−−−√=1350−−−−√ ms/Therefore, the range of the ball (distance from the point of projection) is, time taken by the body to reach the ground is,T=2u sin θg=2×30×sin 4510=32√ sSo, range is,R=Ux×T=1350−−−−√×32√=32700−−−−√=903√ m⇒R=155.88 m
```
one month ago
```							 with the horizontal, and the train is also moving at a speed of 30m/s, hence the total horizontal components of velocities of the ball = 30cosθ + 30 = 30cos45 + 300since the ball is projected at an angle of 45Ux=(30 cos 45)+30 the Time of flight will be 2*30 sin45/10 (taking g = 10m/s2) = 3√2Range = UX  times Time of flight  = (30cos45 +30) * 3√2 =  90(√2+1 )m
```
one month ago
```							Please refer https://www.askiitians.com/forums/Mechanics/a-train-is-moving-on-a-track-at-30m-s-a-ball-is-th_215181.htm
```
one month ago
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