A ball is thrown vertically upwards with a speed of 19.6m/s from the top of a tower, and returns to earth in 6 seconds. Find the height of the tower.

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Solution

Dear Student,

u = 19.6 ms^-1

t = 6 s

Let height of the tower = h

Eqn. of motion of the ball

y = h + ut - ½gt²

Now at t = 6, y = 0

h + 19.6×6 - ½×9.8×62 = 0

=>

thus, height of tower will be

h = 59 m

Regards

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A ball is thrown vertically upwards with a speed of 19.6m/s from the top of a tower, and returns to earth in 6 seconds. Find the height of the tower.

Dear Student, u = 19.6 ms-1 t = 6 s Let height of the tower = h Eqn. of motion of the ball y = h + ut - ½gt2 Now at t = 6, y = 0 h + 19.6×6 - ½×9.8×62 = 0 => thus, height of tower will be h = 59 m Regards

Dear Student,

u = 19.6 ms^-1

t = 6 s

Let height of the tower = h

Eqn. of motion of the ball

y = h + ut - ½gt²

Now at t = 6, y = 0

h + 19.6×6 - ½×9.8×62 = 0

=>

thus, height of tower will be

h = 59 m

Regards