A ball of mass m starts from the top of a ramp at negligible velocity.?

First you need to deteremine the velocity of the mass m when it leaves the ramp:

Using conservation of energy, E(initial)= E(final), then

mgh = ½mv² + mgn => v² = 2g(h-n)

Since the ramp is at a 45° angle, the velocities at the end of the ramp are:

v(x) = v(y) = vsin(45°) = v/sqrt(2) or v(y)² = v²/2 = g(h-n)

Then using equations of motion in the y direction:

v(f)² = v(i)² + 2ay, since v(f) = 0 per requirement that at the landing pad, there will be only horizontal motion, then

v(i)² = -2ay = 2gy or y = v(i)²/2g

So now we substitute our earlier result of v(y)² = g(h-n)

Then y = g(h-n)/2g = (h-n)/2 = (4-2)/2 = 1 m

The height of the landing pad is 1 meter.

Here is what I did:

The speed of the ball when it reaches the end of the ramp is given by: Sqrt(2ghi - 2ghf). Where hi is the initial height, and hf is the final height.

That gave me v = 6.26 m/s to three sig digs.

Since it was at a 45 degree angle, the speed in the y direction, Vy, would equal 6.26*sin(45). That equals 4.43 m/s. The speed in the x direction, Vx, is the same because the sin(45) equals the cos(45).

So, using Vy = Vi - gt, you can show that the ball will reach its maximum height in .452 seconds. At that point, the ball's speed will all be pointing in the x-direction, because the y-component will be zero. I assume that's when you want to the ball to land.

Knowing the time it takes to get there, you can use y = Vi*t - 1/2gt^2, to solve for how high the ball will go. Vi is the 4.43 m/s we found above. Using the time from above, y equals .999 meters.

Also, to find how far away from the ramp to set the landing pad, take the time found above and multiply by Vx, which was 4.43 m/s. That gives you a distance of 1.999 meters.

Of course, all of this is based on no friction! All of these numbers are going to be larger than reality, because friction will come into play.

i bet its 3m (neglecting air resistance) dude! m bit lazy so cant mention the procees ...... first do energy conservation nd then apply basic equations of motion.

hmm, i'm not doing your physics homework for you.