approach:
a) $d_{y} = v_{1y}t + 0.5at^{2}$
$680 = v_{1}\sin{57(6.20s)} - 188.55$
$v_{1} = 585 m/s$
im not sure if this is right.. any help is really appreciated!
B) not sure how exactly to apprach this :/
2026-03-27 05:41:42.1774590102
Kinematics Question?
60 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
The first answer is wrong. The distance $d_{y}$ is negative $=-680$. Then continue the way you have done.
For (B), Take $v_{x}$ the same way you took $v_{y}$. Can you proceed further now?