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An arrow was shot straight up from the ground at time t=0 seconds with an initial velocity of 49 m/sec. Assuming that the free-fall acceleration is -g=-9.8 m/sec^2, determine its maximum height, and when it hits the ground.

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An arrow was shot straight up from the ground at time t=0 seconds with an initial velocity of 49 m/sec. Assuming that the free-fall acceleration is -g=-9.8 m/sec^2, determine its maximum height, and when it hits the ground.

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[tex]v_2^2=v_1^2+2ax[/tex]

Remember that the car will eventually come to rest, so your final velocity will be 0 m/s. For the second part, you can use any other kinematic equation.

The second problem is very similar. At the maximum height, the velocity will be 0 m/s. The same formulas can be used. Just remember that when finding the time to hit the ground, the arrow must travel up and back down.

If you work the problems yourself and want somebody to check if they are right, or if you have more trouble with them, post back with your attempt at the solution and your answers and I will check them for you.

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HallsofIvy

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Similarly, for the second problem, the velocity of the arrow (positive upward) at time t is [itex]v(t)= 49- 9.8t[/itex] and its height is [itex]x(t)= 49t- 4.9t^2[/itex]. Set the second fomula equal to 0 to find the time it hit the ground. set first equation equal to 0 to solve for the time of its highest point (at that point it is no longer going up, so v(t) is not positive but it is not yet going down so v(t) is not negative) and put that t into the equation for x(t) to find the actual height.

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