Question

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An object is thrown vertically downwards from the top of a tower with initial velocity 20 m/s and
reaches the ground in 4 s. The height of the tower is:
(A) 170 m.
(B) 160 m.
(C) 10 m.
(D) 5 m.
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Answer 1

Answer:

B) 160 m

Explanation:

By using,

     s = ut + 1/2 at²

     s = 20 x 4 + 1/2 x 10 x (4)²

     s = 80 + 5 x 16

     s = 80 + 80

     s = 160 m

I hope it will help you      

Answer 2

Answer:

Explanation:

THE HEIGHT OF THE TOWER IS --

Assuming the only force acting on the ball is the initial velocity and standard gravity (9.81 m/s^2): the ball’s position can be written as y=(0.5*a*(t^2))+(v*t)+y1, where ‘a’ is gravity, ‘v’ is the velocity the object is thrown with, ‘y1’ is the initial height in meters, ‘t’ is the time in seconds, and ‘y’ is the height of the object in meters. Using an up as positive reference and plugging in givens, the equation becomes 0=(0.5*(-9.81 m/s^2)*((4 sec)^2))+(v*(4 sec))+(39.2 m). Solving for ‘v’ gives 9.82 m/s.

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