Question

A body is thrown vertically up from the ground with a speed $v_{initial}$ and it reaches maximum height h at time $t=t_o$. What is the height to which it would have risen at time $t=t_o/2$

• A. $0.75h$
  
• B. $h$
  
• C. $0.5h$
  
• D. $0.3h$
  

Answer 1

The correct option is A $0.75h$

In this problem we have free fall motion

$y\left(t\right)=y_{initial}+v_{initial}t-4.9t^2$

$v\left(t\right)=v_{initial}-9.8t$

$v^2\left(t\right)-v_{initial}^2=-19.6\left[y\right(t)-y_{initial}]$

The initial height of the object is $0$ (ground level)

$\therefore y\left(t\right)=v_{initial}t-4.9t^2\to 1$

$v\left(t\right)=v_{initial}-9.8t\to 2$

$v^2\left(t\right)-v_{initial}^2=-19.6\left(t\right)\to 3$

Case I: The object reaches the maximum height h at $t_0$. At the maximum height the velocity of the object is zero and

$y\left(t_0\right)=h$

$h=v_{initial}{t}_0-4.9t_0^2\to 4$

$0=v_{initial}-9.8t_0\to 5$

$v_{initial}^2=19.6h$

Case II: Height of the object at

$t=\frac{t_0}{2}$

Substitute this time in equation $1$ and $3$

$y\left(t_0/2\right)=\frac{v_{initial}{t}_0}{2}=4.9\frac{t_0^2}{4}\to 7$

$v\left(t_0/2\right)=v_{initial}-\frac{9.8t_0}{2}\to 8$

$v^2\left(t_0/2\right)-v_{initial}^2=-19.6y\left(t_0/2\right)\to 9$

We know, $v_{initial}=t_{0}9.8$

Substitute in equation $4$ and $7$

$h=9.8t_0^2-4.9t_0^2=4.9t_0^2\to 10$

$y\left(t_0/2\right)=\frac{9.8t_0^2}{4}=4.9\frac{t_0^2}{4}$

$=3.675t_0^2\to 11$

From equation $10$

$t_0^2=\frac{h}{4.9}$

Substitute this in equation $11$

$y\left(t_0/2\right)=3.675t_0^2$

$=3.675\times \frac{h}{4.9}$

$=0.75h$