Question

The maximum range of a projectile is 500 m. If the particle is thrown up a plane inclined at an angle of 30° with the same speed, the distance covered by it along the inclined plane will be​

Answer 1

Answer:

250m  

500m

750m

100m

Answer :

b

Solution :

For the maximum range, θ=45∘

R=u2sin2θg=u2gsin90∘=u2g

or 500=u2g

The distance covered along the inclined plane can be obtain using the equation

v2−u2=2as

or 0−u2=2(−gsin30∘)s

or s=u2g=500m

Answer 2

Answer:

500

Explanation:

Since, range is maximum, therefore

θ

=

45

∘

.

Hence, using

R

=

u

2

sin

2

θ

/

g

.

We find

500

=

u

2

g

, i.e.

u

=

[

500

g

]

1

2

.

Distance covered along the inclined plane can be obtained using relation

v

2

−

v

2

0

=

2

a

x

⇒

0

−

u

2

=

2

×

(

−

g

sin

30

∘

)

×

x

[

∵

a

=

−

g

sin

30

∘

]

∴

x

=

u

2

g

=

500

m

.