Question

If g is acceleration due to gravity, then the vertical distance travelled by a projectile in time t is equal to.

• A. $\frac{1}{2}g{t}^2$
• B. $g{t}^2$
• C. $\frac{3}{4}g{t}^2$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}g{t}^2$

Initial speed of the projectile in vertical direction is zero i.e. $u_y=0$
Acceleration in the vertical direction $a_y=g$
Vertical distance travelled $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\therefore$ $S_y=0+\frac{1}{2}g{t}^2$
$⟹S_y=\frac{1}{2}g{t}^2$