Answer to Question #287068 in Physics for evan

Question

What is the velocity of a dropped object after it has fallen for 5.0s?

	Formula: v  = v0 +gt  






An arrow is launched straight up with a final velocity of 25.6 m/s. How long before the arrow returned to the ground? 

	Formula:   t = ( v – v0)

			    g






A book bag is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 6 s. With what initial velocity did it leave the ground? (Equation 4)

	Formula: v0  = v -gt






Doug the Talking Dog spots a squirrel.  He accelerates from rest with a constant acceleration of 8 m/s2.  How far will he have moved after 17s? 

	Formula:  x = x0 + v0t + ½ at2




	x =  ½ at2

	x = ½ (8)(17)2

Answer 1

1)

$$v=v_0+gt=0+(-9.8\ \dfrac{m}{s^2})\times5\ s=-49\ \dfrac{m}{s^2}.$$

2)

$$v_{yf}=v_{yi}+gt,$$ $$0=v_{yi}+gt,$$ $$t_{rise}=-\dfrac{v_{yi}}{g}.$$ $$t_{tot}=2t_{rise}=-\dfrac{2v_{yi}}{g}=-\dfrac{2\times25.6\ \dfrac{m}{s}}{-9.8\ \dfrac{m}{s^2}}=5.22\ s.$$

3)

$$v=v_0+gt,$$ $$0=v_0+gt,$$ $$v_0=-gt=-(-9.8\ \dfrac{m}{s^2})\times6\ s=58.8\ \dfrac{m}{s}.$$

4)

$$x=x_0+v_{0}t+\dfrac{1}{2}at^2,$$ $$x=\dfrac{1}{2}at^2,$$ $$x=\dfrac{1}{2}\times8\ \dfrac{m}{s^2}\times(17\ s)^2=1156\ m.$$