Question

A ball is horizontally projected with a speed v from the top of a plane inclined at an angle 45° with the
horizontal. How far from the point of projection with the ball strike the plane?​

Answer 1

Answer:

Writing equation of motion in normal to incline plane:

S

y

=vsinαt−

2

gcosαt

2

For Range , S

y

=0

which gives t=

gcosα

2vsinα

S

x

=vcosαt+

2

gsinαt

2

Replacing the value of t, writing Equation of motion in along incline plane

S

x

=(vcosα)t+(

2

gsinα

)(t

2

)

S

x

=(vcosα)(

gcosα

2Vsinα

)+(

2

gsinα

)(

gcosβ

2Vsinα

2

)

S

x

=

gcosα

2v

2

sinα

(cosα+

cosα

sin

2

α

)

S

x

=

g

2v

2

(

2

1

+

2

1

)=

g

2

2

v

2

solution

Answer 2

Answer:

answer in attachment