The answer is -17x-1. You have to combine like terms.
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
Their velocity afterwards is 2.88 m/s east
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:
where: in this case:
is the mass of the first player
is the initial velocity of the first player (choosing east as positive direction)
is the mass of the second player
is the initial velocity of the second player
is their combined velocity afterwards
Solving for v, we find:
And the sign is positive, so the direction is east.
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