A ditch 2.5 m wide crosses a trail bike path. An upward incline of 15 degrees
has been built up on the approach so that the top of the incline is
level with the top of the ditch. What is the minimum speed a trail
bike must be moving to clear the ditch? (Add 1.4 m to the range for
the back of the bike to clear the ditch safely.)

Homework.

Homework, it really couldn't be more obvious. My answer is do it
yourself, it's the only way you'll learn (I know everyone hates
homework, including me, but getting others to do it for you is not the
way forward).

The Larry Niven & Jerry Pournelle book "Lucifer's Hammer" has a
similar situation. The main character must calculate the speed he has
to drive over a drop to not bottom out the vehicle. Fortunately for
the reader, the character had paid attention in physics class!

hints: A projectile is launched at an angle of 15 Deg. to the
horizontal  with X m/s  and it has to travel (2.5 + 1.4)m horizontal
distance before reaching the ground. Using eq. of motion in two
dimensions find X.
For formul and more explaination Refer this link
<http://www.physicsclassroom.com/Class/vectors/U3L2e.html>

Still If do not get it mail me at nayan.ambali@gmail.com  I will
explain you how to solve it.

$2 is $2.

To do this question you need to split the velocity vector into its
vertical and horizontal components.

Let v=velocity of the bike.

v(sin(15)) = Vertical Velocity

v(cos(15)) = Horizontal Velocity

First, we'll use the vertical velocity to calculate the time spent in the air.

x(t) = (-1/2)(g)(t^2) + v(0)t + x(0)

We know when the bike hits the ground on the other side x(t) will be
zero.  g is the accelleration due to gravity (about 9.8).

0 = -1/2(g)(t^2) + v(sin(15))t

1/2(g)(t^2) = v(sin)(15)t

Divide by t.  (this eliminates the solution t=0 because we can't div
by 0.  its wrong anyway because t=0 is the start of the jump)

1/2(g)(t) = v(sin 15)
t = 2v(sin 15)/g

Now time for the horizontal part.  The velocity is constant at v(cos
15), so the distance travelled(length of ditch = 2.5m) is v(cos 15)(t)
(velocity x time).

2.5 = v(cos 15)(t)

plug in for t from the above equation:

2.5 = v(cos 15)2v(sin 15)/(g)

v^2 = (2.5g)/(2(cos 15)(sin 15))

If you remember trig identities:
sin(2a) = 2 sin(a) cos(a)
sin(30) = 2 sin(15) cos(15)

v^2 = 2.5g/sin (30)

sin(30) = 1/2

v^2 = 2.5g/(1/2) = 5g * 49

v = 7m/s

Obsidianfang, 7m/second is about 25 kph or a a bit less than 16 mph.
Sounds a bit low to clear an 8 foot wide trench, doesn't it?

You forgot the extra 1.4 meters the problem called for. Adding that in
gives 19.5 m/second or 70 kph, roughly 45 mph.