Answer to Question #162165 in Classical Mechanics for pat

Question

A stone is thrown upward with a velocity of 4.0 m/s dropped from a bridge that is 15 metres above the water.

(a) How high above the bridge does the stone rise?(a) How high above the bridge does the stone rise?

(b) How long will it take for the stone to reach the water?

(c) What will be the velocity of the stone when it hits the water?

Answer 1

$\text{ Acceleration of gravityand body speed } \newline \text{are directed in different directions}\newline V= V_0-gt\newline h =V_0t-\frac{gt^2}{2}\newline a)h_{max} \text { if } V=0\newline V_0-gt=0\newline t=\frac{V_0}{g}\newline h_{max} =V_0t-\frac{gt^2}{2}=\frac{V_0^2}{2g}\newline h_{max} =\frac{4^2}{2*9.8}\approx0.82\newline$

$b) h = -15$

$V_0t-\frac{gt^2}{2}=-15$

$\frac{9.8*t^2}{2}-4t-15 =0$

$t_1\approx2.23\ t_2\approx-1.4(\text{not correct});$

$t =2.23$

$c)V=V_0-gt$

$V=4-9.8*2.23\approx-17.85$

$|V|=17.85$

Answer:a)0.82 m b)2.23 s c)17.85 m/s