Question

Two balls A and B of masses 25 g and 50 g
respectively are thrown horizontally in same direction
with the speed of 20 m/s and 25 mis respectively at
the same time from the top of a building of height
125 m. The separation between the balls when they
reach the ground is (g = 10 m/s2)
(1) 10 m
(2) 20 m
(3) 15 m
(4) 25 m​

Answer 1

Answer:Answers

THE BRAINLIEST ANSWER!

abhi178

Abhi178The Sage

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

Explanation:

Answer 2

Answer:

Opt 4 - 25m

Explanation: We have to use the horrizontal projectile motion formula which is y = gx^2/ 2u^2

here we have to find x in both cases i.e, for the ball with 20m/s vel and for the ball with 25m/s vel.

X=$\sqrt{y*2u^{2}/g }$

case i X = 100  

case 2 X=125

125-100= 25m