Question

A ball is thrown upward from the top of a $31.0m$ tower on an unknown planet (gravitational acceleration is unknown) with an initial speed of $12.0m/s$ and hits the ground with a speed of $52.0m/s$. How long was the ball in flight? (in s)

• A. $1.15$
  
• B. $2$
  
• C. $1.55$
  
• D. $1.6$
  

Answer 1

The correct option is C $1.55$

This is the free fall motion

$\therefore y\left(t\right)=y_{initial}+v_{initial}-\frac{g{t}^2}{2}$

$v\left(t\right)=v_{initial}-gt$

$v^2\left(t\right)-v_{initial}^2=-2g[y\left(t\right)-y_{initial}]$

where-g is unknown acceleration

(it is negative-pointing downward, then quantity g is positive)

Initial height$:y_{initial}=31m$

Initial speed (it is pointing upward which means that the y-component of velocity is positive):-$v_{initial}=12㎧$

The final speed is negative-pointing downward

$v\left(t\right)=-52㎧$

The final height is $0$

By substituting this values in above equation

$0=31+12t-\frac{g{t}^2}{2}$

$-52=12-gt$

${52}^2-{12}^2=-2g(0-31)$

From last equation, unknown acceleration is

$g=41.3㎨$

By substituting this value in second equation

$t=\frac{52+2}{g}=\frac{52+12}{41.3}$

$\therefore t=1.55s$