Question

At time t=0.0s, a steel ball is thrown vertically downwards from the top of a tall tower with a velocity of 2.0 m s−1. Calculate the distance it falls between the period of t=2.0 and t=3.0 s. (You may ignore air resistance. Take the acceleration due to gravity to be 10 m s−2.)

Answer 1

Answer:

here u=2m/s

t=2.0sec

s=?

so,

s=ut+1/2at^2

=2×2+1/2×10×[2]^2

=4+5×4

=24 m. ans

t=3.0sec

so,

s=ut+1/2at^2

=2×2+1/2×10×(3)^2

=4+5×9

=49 m ans

so, here is your answer

hope it helps you

Answer 2

Answer:

27 m

Explanation:

Y= Vi(t)+ 1/2a(t)^2

t= 2.0 s

Y= 2.0(2.0)+ 1/2(10)(2)^2

Y= 24 m

t= 3.0 s

Y= 2.0(3.0)+ 1/2(10)(3)^2

Y= 51 m

Y=51 - 24

Y= 27 m