Question

Two balls are thrown horizontally from a 80 m high tower in same direction with velocities 2m / s and 3m / s respectively. The separation between the two balls when they hit the ground is
(1)4 m (2)8 m (3)12 m (4)20m​

Answer 1

Answer:

Seperation of balls when they hit ground = (1) 4m

Explanation:

Two balls are thrown horizontally,

Thus their initial vertical velocity will be zero.

Both balls will travel same vertical distance S = 80m

Thus, time taken by the balls to hit the ground, can be calculate by

S = ut + (1/2)at²

where - a = 10 m/s² (acceleretion due to gravity)

∴ 80 = 0×t + (1/2)× 10 × t²

∴ t² = 80 × 2 ÷ 10

∴ t² = 16

∴ t = √16

∴ t = 4 s

In this time both ball will travel their horizontal distances.

∴ For Ball 1 - Distance travelled = Speed × time

∴ D1 = 2  × 4

∴ D1 = 8 m

For ball 2 - Distance travelled D2 = Speed × time

∴ D2 = 3 × 4

∴ D2 = 12 m

∴ Seperation of balls when they hit ground = D2 - D1 = 12 - 8 = 4 m

∴ Seperation of balls when they hit ground = (1) 4m

Answer 2

answer : 4m  

explanation : Let’s solve with different method.  

     here it is clear that balls are thrown horizontally so, the path of balls are parabolic.

 it is example of horizontal projectile.  equation of path of horizontal projectile is y = gx²/2u²

where y denoted vertical distance e.g., height of ball from which it is thrown. x is horizontal distance, g is acceleration due to gravity and u is initial velocity of body.  

horizontal distance covered by first ball , x = $\sqrt{\frac{2u^2y}{g}}=\sqrt{\frac{2(2)^280}{10}}=8m$  

horizontal distance covered by 2nd ball, x’ = $\sqrt{\frac{2u^2y}{g} } =\sqrt{\frac{2(3)^280}{10}}=12m$

 hence, The separation between the two balls when they hit the ground is (12m - 8m) = 4m