Question

A projectile is fired upward at an angle of 30 with a velocity of 40m/s the time taken to reach the ground by projectile after instant of firing is

Answer 1

Answer:

• The time taken (T) by the projectile is 4 seconds.

Given:

1. Initial velocity (u) = 40 m/s.
2. angle of projection (θ) = 30°
3. Acceleration due o gravity (g) = 10 m/s²

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\large \bigstar\; \boxed{\tt T = \dfrac{2\;u\; \sin \theta}{g}}$

$\frak{Here}\begin{cases}\text{T Denotes Time period}\\\text{u denotes Initial velocity}\\\text{ \theta Denotes angle of projection}\end{cases}$

Now,

$\large\boxed{\tt T = \dfrac{2\;u\; \sin \theta}{g}}$

Substituting the values,

$\displaystyle\dashrightarrow\tt T=\dfrac{2\times40\times \sin 30^\circ}{10}\\\\\\\dashrightarrow\tt T=\dfrac{80\times \sin 30^\circ}{10}\\\\\\\dashrightarrow\tt T=8\times\sin 30^\circ\\\\\\\dashrightarrow\tt T=8\times\dfrac{1}{2}\\\\\\\dag\; \sin30^\circ=\dfrac{1}{2}\\\\\\\dashrightarrow\tt T=\dfrac{8}{2}\\\\\\\dashrightarrow\tt T=\cancel{\dfrac{8}{2}}\\\\\\\dashrightarrow \large {\underline{\boxed{\red{\tt T=4\;sec}}}}$

∴ The time taken (T) by the projectile is 4 seconds.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

$\boxed{\begin{minipage}{7cm}\underline{\bf{Important Formulas}}\colon\\\\\bullet\;\tt H_{max}=\dfrac{u^2\sin^2\theta}{2\;g}\\\\\bullet\;\tt R=\dfrac{u^2\sin 2\;\theta}{g}\\\\\bullet\;\tt H_{max}=\dfrac{g\;t^2}{8}\\\\\bullet\;\tt R\tan\theta=4\;H\end{minipage}}$

$\rule{300}{1.5}$

Answer 2

AnswEr :

Explanation :

From the Question,

• Angle of Projection (∅) = 30°

• Net initial velocity (u) = 40 m/s

We have to find out the time taken by the projectile to reach the ground after firing

Time of Flight

$\huge{\boxed{\boxed{\sf T = \dfrac{2u sin \theta}{g}}}}$

(Putting the values)

$\longrightarrow \sf T = \dfrac{2 \times 40}{10 \times 2} \\ \\ \large{\longrightarrow \fbox{\sf T = 4s}}$

The projectile would take 4s to reach the ground after firing