If I throw a ball straight up with an initial speed of 35 m/s, how long will it take to return to me?
2007-07-06 04:42:41 · 4 answers · asked by Chablisah 1 in Science & Mathematics ➔ Physics
You can use the equation of motion:
a= (v2-v1)/t
And solve for t:
t=(v2-v1)/a
v1 is 35m/s, and v2 is -35m/s because (neglecting air resistance) if an object is thrown upward with some velocity v, it comes down and at the same height at which it was thrown, has a velocity of -v.
a is the acceleration due to gravity, which is -9.8m/s^2.
Note the negative, because the acceleration is a downward acceleration. Plug in the number and there you go.
Hope this helps.
2007-07-06 04:45:30 · answer #1 · answered by Mikey C 2 · 1⤊ 0⤋
v = u - gt; where v = 0, the velocity at the top of the ball's travel, u = 35 m/sec the initial velocity out of your hand, g = 9.81 m/sec^2 to acceleration due to gravity on Earth's surface, and t = 1/2 T where t = the time to reach apex and v = 0 and T is the total time up and back down into your hand (the answer you're looking for).
Therefore, we have 0 = u - gt; so that u = gt = g(1/2 T) and 2u/g = T and you can do the math.
You should spend some time memorizing and understanding the physics behind the SUVAT equations of motion (v = u - gt is one of them). Browse "SUVAT" and you'll find lots of web sites explaining these equations, which you will see many many times in introductory physics classes.
2007-07-06 05:18:53 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
3 hours, 14 minutes due to Hasslein's theory of time and speeds continuums.
2007-07-06 04:56:59 · answer #3 · answered by tracyterry 3 · 0⤊ 2⤋
it depends on the elasticity of the ball...you should state the time taken
2007-07-06 04:47:13 · answer #4 · answered by rudz 2 · 0⤊ 2⤋