Question

Two balls are dropped from different heights at different instants. Second ball is
dropped 2 seconds after the first ball. If both balls reach the ground simultaneously
after 5 seconds of dropping the first ball, then the difference between the initial
heights of the two balls will be. (g = 9.8m/s).
a) 58.8m ub) 78.4 m (c) 98.0m (d) 117.6 m​

Answer 1

Answer:

Explanation:

For first ball S=ut+  

2

1

​  

gt  

2

∴(t=5sec)

=0+  

2

1

​  

g(5)  

2

=  

2

25g

​  

=H  

1

​  

 

For second ball

S=ut+  

2

1

​  

gt  

2

∴(t=5−2=3sec)

=0+  

2

1

​  

g(3)  

2

=  

2

9g

​  

=H  

2

​  

 

Difference between the initial heights

H  

1

​  

−H  

2

​  

=  

2

25g

​  

−  

2

9g

​  

 

=  

2

16g

​  

 

=8g=8×9.8

=78.4

∴ The difference of initial heights of the two balls is 78.4m