Question

A projectile is thrown from the base of an incline of angle 300 as shown in the figure. It is thrown at an angle 600 from the horizontal direction at a speed of 10 m/s. The total time of flight is (consider g  10 m / s2 ) : (Upto two decimal places)

Answer 1

Correct Question:

A projectile is thrown from the base of an inclined plane of angle 30°. It is thrown at an angle 60° from the horizontal direction at a speed of 10 m/s. The total time of flight is (consider g=10 m/s²): (Upto two decimal places)

Given:

Initial speed of projectile,u= 10 m/s

Angle between inclined plane and horizontal direction,θ= 30°

Angle between direction of velocity and horizontal direction, β= 60°

To Find:

Total time of flight of projectile

Solution:

We know that,

• Time of flight T of projectile on inclined plane is given by

$\pink{\boxed{\bf{T=\dfrac{2u\:sin\alpha}{g\:cos\theta}}}}$

where,

u is initial velocity of projectile

α is angle between direction of velocity and inclined plane

θ is angle between inclined plane and horizontal direction

$\rule{190}{1}$

Let the angle between direction of velocity and inclined plane be α

So, according to the question

$\longrightarrow\rm{\alpha=\beta-\theta}$

$\longrightarrow\rm{\alpha=60^{\circ}-30^{\circ}}$

$\longrightarrow\rm{\alpha=30^{\circ}}$

$\rule{190}{1}$

Now, let the total time of flight of projectile be T

So,

$\longrightarrow\rm{T=\dfrac{2u\:sin\alpha}{g\:cos\theta}}$

$\longrightarrow\rm{T=\dfrac{2\times10\times sin30^{\circ}}{10\times cos30^{\circ}}}$

$\longrightarrow\rm{T=\dfrac{\cancel{20}\times\dfrac{1}{\cancel{2}}}{\cancel{10}\times\dfrac{\sqrt{3}}{\cancel{2}}}}$

$\longrightarrow\rm{T=\dfrac{\cancel{10}}{\cancel{5}\sqrt{3}}}$

$\longrightarrow\rm{T=\dfrac{2}{\sqrt{3}}}$

$\longrightarrow\rm\green{T=1.15\ s\ (approx.)}$

Hence, the total time of flight of projectile is 1.15 s.