Question

* An object is dropped from the top of the building
with the constant acceleration of 5m\s^2? find its speed
after 5 sec after it was dropped.​

Answer 1

Answer: Distance covered by first ball 0+  

2

1

​  

×10×3  

2

=45

Distance covered by second ball 5×3+  

2

1

​  

×10×3  

3

=15+45=60

difference in height =15 m

Answer 2

Answer:

its 25

Explanation:

we know accleration=v-u/t

so we cross multiply and (5*5) which is 25

(v-u/t=a)5m/s^2=0m/s-v/5

cross multiply and we get 25 thats all :)