Answer to Question #103551 in Mechanics | Relativity for nylasha cabness

Question

you throw a ball straight up with an initial velocity of 15.0m/s letting it go at 1.5m above the ground it passes a tree branch on the way up at a height of 7.00m how much additional time will pass before the ball passes the tree branch on the way back down?

what kinematics equations do you plan to use to solve this problem?
at what times does it pass the branch the first time on the way up ?
at what time does it pass the branch the second time on the way down?
what is the answer to the question asked on the problem?

Answer 1

$$v_A^2=v_O^2+2gy$$

$$v_A=\sqrt{15^2-2(9.8)(7)}=9.37\frac{m}{s}$$

$$v_B=v_A+gt_{AB}\to 9.37=15-(9.8)t_{AB}$$

$$t_{AB}=0.956\ s$$

$$t_{AA'}=2t_{AB}=2(0.956)=1.91\ s$$