A typical ramp is 65 degrees. A skier attains a height of 11m above the ramp. what is the launch speed?

Question

This is a two dimensional kinematics problem from physics. This physics is not calculus based. The question is asking for the skier's launch speed as she jumps off of a 65 degree ramp. In other words I need to find out the horizontal velocity.

Answer 1

Since the skier leaves the ramp at an angle of 65 deg, there is a vertical component and a horizontal component. The vertical component is decelerated by gravity, so the maximum height reached (when the vertical component is zero) can be used to calculate the initial velocity.

Deal with the vertical component first:

The distance which a free-falling object has fallen from a position of rest is also dependent upon the time of fall. The distance fallen after a time of t seconds is given by the formula below:

d = 0.5 * g * t^2

so t = sqrt( d / (0.5 * 9.8))

we know that d = 11 so t = 1.5

Knowing the time taken, we can now calculate the velocity:

The velocity of a free-falling object which has been dropped from a position of rest is dependent upon the length of time for which it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is:

Final Velocity = g * t

where g is the acceleration of gravity (approximately 10 m/s/s on Earth; its exact value is 9.8 m/s/s). The equation above can be used to calculate the velocity of the object after a given amount of time.

So the Initial Vertical Velocity after 1.5 seconds is:

9.8 * 1.5 = 14.7m/s

This is the vertical component. Since the angle is 65 degrees, the actual velocity is:

14.7 / Sin 65 = 16.2 m/s

and the horizontal component will be:

16.2 * Cos 65 = 6.8 m/s

Answer 2

This is simply a mgh = 1/2mv^2 question if you disregard air friction

Answer 3

maximum height reached= (v^2/2g)sin@

hence launch velocity=15.4 m/s
horizontal velocity= vcos@=6.5 m/s

Answer 4

150 MPH

Answer 5

TAKEING A BONG HIT TELL YA LATER WOW DUDE THATS FAST.