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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.

Kevin Nash
6 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.
Tanya
23 Points
3 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°
Rishab
21 Points
3 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...