Question

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

Answer 1

Answer:

59m

Step-by-step explanation:

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/s2

Now, the height is

From equation of motion

  s=ut+21at2

 s=19.6×6−21×3.26×36

 s=58.9

 s=59m

Hence, the height of the tower is 59 m

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Answer 2

Answer:

Taking upward direction to be positive

u=19.6m/s

t=6s

a=-g=-9.8m/s^2

Displacement of the ball=-h

Applying s=ut+(at^2)/2

-h=19.6(6)+(-9.8)(36)/2

h=58.8m