Stacy Heath

2023-02-18

A ball thrown vertically upwards with a speed of $20\text{}m{s}^{-1}$ from the top of a tower reaches the earth in 8 s. Find the height of the tower. Take $g=10\text{}m{s}^{-1}$

Destiney Manning

Beginner2023-02-19Added 4 answers

Let the height of the tower be h. Assume that things are moving in the right direction.

The height of the tower determines the displacement's size. Compared to the initial position, the final position (ground) is downward (or negative).

$\therefore s=-h$

Acceleration is in downwards direction.

$\therefore a=-g=-10\text{}m{s}^{-2}$

From the second equation of motion,

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow -h=ut-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow -h=20\times 8-(\frac{1}{2}\times 10\times {8}^{2})\phantom{\rule{0ex}{0ex}}\Rightarrow h=160\text{}m$

The height of the tower determines the displacement's size. Compared to the initial position, the final position (ground) is downward (or negative).

$\therefore s=-h$

Acceleration is in downwards direction.

$\therefore a=-g=-10\text{}m{s}^{-2}$

From the second equation of motion,

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow -h=ut-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow -h=20\times 8-(\frac{1}{2}\times 10\times {8}^{2})\phantom{\rule{0ex}{0ex}}\Rightarrow h=160\text{}m$