Question

a body of mass 2kg has initial velocity 3 m/s along horizontal and it is subjected to a force of 4 newtons in vertical direction perpendicularly. the distance of body after 4 seconds ​

Answer 1

Answer:

28 m

Explanation:

m=2 kg

u= 3m/s

f=4N

d=? (after 4 sec)

we know that: f= ma

so 4 =2 a

a= 4/2=2m/s^2

by using

s= ut+1/2at^2

s= 3*4 + 1/2 (2)4^2

s= 12+ 16

s = 28m