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A projectile is thrown from the base of an incline of angle 30° as shown in the figure. It is thrown at an angle of 60° from the horizontal direction at a speed of 10 m/s. The total time of flight is (consider g = 10 m / s 2 ) :

Projectile motion on an incline

Projectile motion on an incline

a 2 b 3 c 3 2 d 2 3

This situation can be handled with a reoriented coordinate system as shown in the figure. Here, angle of projection with respect to x - direction is (θ-α) and acceleration in y - direction is "g cosα". Now, total time of flight for projectile motion, when points of projection and return are on same level, is :

Projectile motion on an incline

Projectile motion on an incline

T = 2 u sin θ g

Replacing "θ" by "(θ-α)" and "g" by "gcosα", we have formula of time of flight over the incline :

T = 2 u sin θ - α g cos α

Now, θ = 60° , α = 30°, u = 10 m/s. Putting these values,

T = 2 X 10 sin 60 0 - 30 0 g cos 30 0 = 20 sin 30 0 10 cos 30 0 = 2 3

Hence, option (d) is correct.

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Two projectiles are thrown with the same speed from point "O" and "A" so that they hit the incline. If t O and t A be the time of flight in two cases, then :

Projectile motion on an incline

Projectile motion on an incline

a t O = t A b t O < t A c t O > t A d t O = t A = u tan θ g

We have discussed that projectile motion on an incline surface can be rendered equivalent to projectile motion on plane surface by reorienting coordinate system as shown here :

Projectile motion on an incline

Projectile motion on an incline

In this reoriented coordinate system, we need to consider component of acceleration due to gravity along y-direction. Now, time of flight is given by :

t = 2 u y g cos θ

Let us first consider the projectile thrown from point "O". Considering the angle the velocity vector makes with the horizontal, the time of flight is given as :

t O = 2 u sin 2 θ θ g cos θ t O = 2 u tan θ g

For the projectile thrown from point "A", the angle with horizontal is zero. Hence, the time of flight is :

t A = 2 u sin 2 X 0 + θ g cos θ = 2 u tan θ g

Thus, we see that times of flight in the two cases are equal.

t A = t O

Hence, option (a) is correct.

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A ball is projected on an incline of 30° from its base with a speed 20 m/s, making an angle 60° from the horizontal. The magnitude of the component of velocity, perpendicular to the incline, at the time ball hits the incline is :

Projectile motion on an incline

Projectile motion on an incline

a 10 m / s b 10 3 m / s c 20 3 m / s d 20 3 m / s

The velocity in y-direction can be determined making use of the fact that a ball under constant acceleration like gravity returns to the ground with the same speed, but inverted direction. The component of velocity in y-direction at the end of the journey, in this case, is :

Projectile motion on an incline

Projectile motion on an incline

v y = u y = - 20 sin 30 0 = - 20 X 1 2 = - 10 m / s

| v y | = 10 m / s

Hence, option (a) is correct.

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A projectile is projected from the foot of an incline of angle 30°. What should be the angle of projection, as measured from the horizontal direction so that range on the incline is maximum?

Projectile motion on an incline

Projectile motion on an incline

a 45 ° b 60 ° c 75 ° d 90 °

We can interpret the equation obtained for the range of projectile :

R = u 2 g cos 2 α [ sin 2 θ - α sin α ]

The range is maximum for the maximum value of "sin(2θ-α) “ :

sin 2 θ α = 1 = sin 90 0 2 θ 30 0 = 90 0 θ = 60 0

Hence, option (b) is correct.

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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