Can you help me with this Physics problem
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Can you help me with this Physics problem

[From: ] [author: ] [Date: 11-09-27] [Hit: ]
6 seconds from when you threw it to when you caught it, how high did it go?Time up is equal to time down so it took 2.8 seconds to reach its zenith, correct? All the while its fighting the negative acceleration of gravity which is -9.......
Here it is:

"You toss a ball straight up in the air, it goes up, comes down, and you catch it. If it took 5.6 seconds from when you threw it to when you caught it, how high did it go?"

Time up is equal to time down so it took 2.8 seconds to reach it's zenith, correct? All the while it's fighting the negative acceleration of gravity which is -9.8 m/s^2. So 9.8 x 2.8 = ~27.44 meters high, right?

I think that's the right answer but what motion equation(s) do you use to get to that answer?

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Set up the problem like this

Equation of motion
h = h0 + v0*t - 1/2*g*t^2
(h - h0) = v0*t - 1/2*g*t^2

If it starts and ends at the same height then

(h - h0) = 0

0 = v0*t - 1/2*g*t^2

Solve for v0

0 = v0 - 1/2*g*t

v0 = 1/2*g*t = 1/2*9.81 m/s^2 * 5.6 s = 27.45 m/s

At the apex of the throw, the velocity in the vertical direction = 0 m/s
Use vf^2 = v0^2 - 2*g*h

0 = (27.45 m/s)^2 - 2 * 9.81 m/s^2 * h

h = 38.46 m

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No do d(t)= at^2 +v at 0 (t) + d at zero plug in ur number to solve for v and calculate in this equation whatever u want
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