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`         A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground. `
5 years ago

```							Sol. It is a case of projectile fired horizontally from a height.
h = 100 m, g = 9.8 m/s2
a) Time taken to reach the ground t = √((2h/g) )
= √((2 x 100)/9.8) = 4.51 sec.
b) Horizontal range x = ut = 20 × 4.5 = 90 m.
c) Horizontal velocity remains constant through out the motion.
At A, V = 20 m/s
A Vy = u + at = 0 + 9.8 × 4.5 = 44.1 m/s.
Resultant velocity Vr = √((44.1)^2  + 〖20〗^2 ) = 48.42 m/s.
an β = v_y/V_x  = 44.1/20  2.205
⇒β = tan–1 (2.205) = 60°.
The ball strikes the ground with a velocity 48.42 m/s at an angle 66° with horizontal.

```
5 years ago
```							It is the case of projectile from height     Speed = 20m/s    Height = 100m       Time = ut + 1/2 at^2        100  = 0t + 1/2 * 10 * t^2          t^2 = 100/5            t   = 4.5 secondsDistance = speed × time              s = 20 × 4.5              s = 90 m Horizontal velocity remains constant through out the motion. At A, V = 20 m/s A Vy = u + at = 0 + 9.8 × 4.5 = 44.1 m/s. Resultant velocity Vr = √((44.1)^2 + 〖20〗^2 ) = 48.42 m/s. an β = v_y/V_x = 44.1/20 2.205 ⇒β = tan–1 (2.205) = 60°. Therefore the angle is 60°
```
2 years ago
```							As we are given that H=100m and u=20m/s so we know that it is a projectile question firstlya)  s= ut + 1/2at^2 so      100=1/2 (10)t^2       ( as u=0 in vertical component) So we get t= 2√5 I.e 4.5 sec. b) again s= ut + 1/2 at^2      So S=ut       (as a=0 in horizontal component no acceleration)   S=20×2√5  I.e 90m. Thank-you...
```
2 years ago
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