Question

The height and horizontal distance $x$ of a projectile are given by $y=(8t-5t^2)m$ and $x=6tm$ where $t$ is in seconds. Find the maximum height of a projectile.
(Take $g=10m/s^2$)

• A. $4.7m$
• B. $5.6m$
• C. $3.2m$
• D. $8.7m$

Answer 1

The correct option is C $3.2m$

Compare the equation $y=8t-5t^{2}m$ from first equation of motion in $y$ - direction,
$y=u_{y}t+\frac{1}{2}g{t}^2$
$\Rightarrow u_y=8m/s$
Since maximum height reached, $H=\frac{u_y^2{\sin }^2\theta }{2g}$
We know maximum height is achieved when $\theta ={90}^o$
$H=\frac{u_y^2}{2g}=\frac{8^2}{20}=3.2m$