Question

A body is projected from the top of a tower of height $32\text{m}$ with a velocity of $20\text{m/s}$ at an angle of ${37}^{\circ }$ above the horizontal. The time of flight of the projectile is
(Take $g=10m/s^2$)

• A. $\frac{8}{5}\text{s}$
• B. $2\text{s}$
• C. $8\text{s}$
• D. $4\text{s}$

Answer 1

The correct option is D $4\text{s}$

Time of flight only depends on the vertical component of velocity.
Initial vertical velocity $u_y=20\sin {37}^{\circ }=20\times \frac{3}{5}=12m/s$ (taking upward direction as positive)

Displacement of the body when it reaches the ground $h=-32\text{m}$

Using the second equation of motion,
$h=u_{y}t-\frac{1}{2}g{t}^2$
$\Rightarrow -32=12t-5t^2$
$\Rightarrow 5t^2-12t-32=0$
$\Rightarrow 5t^2-20t+8t-32=0$
$\Rightarrow 5t(t-4)+8(t-4)=0$
$\Rightarrow (5t+8)(t-4)=0$
$\Rightarrow t=4$ (since time cannot be negative)
Hence the time of flight is $4\text{s}$.