Question

a body is projected vertically up. what is the distance covered in its last second of upward motion​

Answer 1

Answer:

S=(u+v)t÷2

Step-by-step explanation:

u=initial velocity

v=final velocity

t=time

S is the distance covered

Answer 2

Answer:

$\huge\mathsf\pink{Solution:-}$

→The distance covered by the object in its last second of its upward motion its equal to the distance covered in the first second of its downward motion.

Hence, s = ½ gt² = ½ × 10 × 1 = 5m