Conservation of Energy

Question

A 45 kg student runs with a speed of 2.0 m/s off a horizontal ledge that is 5.0 above a swimming pool. Calculate the vertical speed of the student when she hits the water.

 

...

 

Energy before = Energy after

mgh + (m*v^2)/2 = mgh + (m*v^2)/2

(45*9.8*5) + (45*2^2)/2 = 0 + (45*v^2)/2

4590 = 45v^2

v= 10.1 m/s

 

Ans: 9.9 m/s ??

 

 

Answer 1

Gnarls,

 

You correctly computed the TOTAL speed when it hits the water, but the problem asked for the vertical speed only.  There is more than one way to get this.  The tutors gave you correct ways to get the vertical speed.  Another way, using your value for v of 10.1 m/s,

 

v² = v_x² + v_y²

 

v = 10.1

v_x = 2.0

 

10.1² = 2.0² + v_y² 

 

v_y = √(10.1² - 2.0²) = 9.9

 

Answer 2

The vertical speed can be computed from the conversion of the potential energy to kinetic energy, mgh = (1/2)mv²

 

√(2gh) = v

 

The horizontal speed, 2.0 m/s, remains unchanged because there is no force acting in the horizontal direction.

Answer 3

Hi Gnarls,

Your answer of 10.1 m/s is correct if you are asking for the velocity with which the student hits the water.  However, you are asked for the VERTICAL velocity.  The 10.1 m/s you calculated is NOT the vertical velocity, rather it is the vector sum of the horizontal and vertical components.  

 

One way to get the vertical component of the velocity is to isolate the vertical components of motion and use the equation v_f² = v_i² + 2ad.  

 

v_f² = 0 + 2(9.8)(5)

v_f²= 98

v_f = 9.9

 

Hope this helps!  :)