Question

A ball thrown vertically upwards with a speed of 19.6 m/s from the top of a tower returns to the base of tower in 6s. Find the height of tower : ( Assume accelaration due to gravity $9.8m/s^2$)

• A. 60 m
• B. 100 m
• C. 58.8 m
• D. None of these

Answer 1

The correct option is C

58.8 m

$s=ut+\frac{1}{2}a{t}^2$

Let the height of the tower be $h$.

Here,

$u=19.6m/s$

$t=6s$

$s=-h$ (The magnitude of the displacement is the height of the tower. Since it is in the direction opposite to the initial velocity, its negative value is taken).

$a=-g=-9.8m/s^2$ (It is taken as negative because the direction of acceleration due to gravity is downwards, while the initial velocity is upwards).

$\therefore -h=(19.6\times 6)-\frac{9.8\times 6\times 6}{2}$

$=-58.8m$

Thus, the height of the tower is 58.8 m.