Question

4 A ball is thrown vertically upwards
with a speed of 19.6 m/s from
the top a
tower. It returns to
the to the earth in 6s. Find the height
of the tower?​

Answer 1

Answer:

Given that,

Initial velocity u=19.6m/s

Final velocity v=0

Time t=6sec

The acceleration is

We know that,

v=u+at

0=19.6+a×6

a=−3.26m/s(square)

Now, the height is

From equation of motion

s=ut+ 1/2 at(square)

s=19.6×6− 1/2x3.26x36

s=58.9

s=59m

Hence, the height of the tower is 59 m