Using energy considerations to determine speed

[I_Try_Math]

Homework Statement

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water with an initial speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent of the direction thrown. (Hint: show that $$K_i + U_i = K_f + U_f$$)

Relevant Equations

$$K_i + U_i = K_f + U_f$$

$$K_i + U_i = K_f + U_f$$

$$K_i = \frac 1 2 m(15)^2$$

$$U_i = 196m$$

$$U_f = 0$$

$$K_f = K_i + U_i - U_f$$

$$=\frac {15^2} 2 m + 196m$$


$$=\frac 1 2(15^2m + 98m)$$

$$=\frac 1 2m(15^2 + 98)$$

$$=\frac 1 2m(323)$$

$$=\frac 1 2m(17.97)^2$$

$$v=17.97 m/s$$


Not seeing where I'm making a mistake. Any help is appreciated.

 

[PeroK]

$$=\frac {15^2} 2 m + 196m$$
$$=\frac 1 2(15^2m + 98m)$$

?

 

[I_Try_Math]

?

Well...that wasn't very smart. Anyway thanks for pointing that out.

 

[PeroK]

PS you could have noted that:
$$E_i = E_f \ \Rightarrow \ \frac 1 2 mu^2 + mgh = \frac 1 2 mv^2$$$$\Rightarrow \ v^2 = u^2 + 2gh$$Which should look familiar.

 

[Orodruin]

As a side note: Never insert numbers into equations unnecessarily. In particular not if you are going to remove the units. Only insert numbers once you reach a final result.