+60 You toss a ball straight up in the air, it goes up, comes down, and you catch it. If it took 3.5 s from when you threw it to when you caught it, how high did it go?
There is no air resistance, and acceleration due to gravity is 9.8 m/s.
I need to know the steps to solve this problem, so that I'll know how to solve similar problems in the future.
Any help is apreciated, thanks!
+36915 it goes up in 1/2 of the total time....it comes down in the other half of the time.
1/2 of the time is 1.75 secs
vf=vi+at to find vi (velocity initial) vf= 0 (because it stops and begins to fall back down)
then use
Use xf=xi + vt + 1/2 at^2 to find xf (the final height) with xi = initial height = 0
vf=0 = vi + (-9.8)( 1.75) yields vi = 17.15 m/s
xf = xi + 17.5(1.75) + 1/2(-9.8)(1.75)^2
xf= 0 + 30.625 - 15.006= 15.619 m
+223 So there are 5 different values that could be used to solve kinematic problems like this one.
In this case, there is:
1) initial velocity (v0) - we are not provided with this value
2) final velocity (vf) - we are not provided with this value
3) acceleration (a) - this is provided in the problem as -9.8 m/s2
4) time (t) - this is 3.5 seconds
5) delta x (in other words, the displacement) - we are not provided with this value
The kinematic equation we will be using is:
delta x = v0t + (1/2)at2
If we plug in the values we know, we cannot solve the equation for delta x since there will still be two variables. So we need to find what the initial velocity is by using this equation:
vf = v0 + at
But in order to find the initial velocity, we need to find the final velocity, which we also don't know. But, we do know that the velocity is 0 at the height of the toss, and we also know that at the height of the toss, half the time has elapsed (half of 3.5 is 1.75). So we end up with.
0 = v0 + (-9.8 m/s2)(1.75 s) ----> 0 = v0 + -17.15 m/s
Then we isolate the v0 on the left side of the equation.
v0 = 17.15 m/s
Now we can plug this into the equation. (Remember to plug in the time as half of the given time)
delta x = v0t + (1/2)at2 ----> delta x = (17.15 m/s)(1.75 s) + (1/2)(-9.8 m/s2)(1.75 s)2
Simplify the equation to get:
delta x = 15 m
The ball reaches a height of 15 meters.
+36915 it goes up in 1/2 of the total time....it comes down in the other half of the time.
1/2 of the time is 1.75 secs
vf=vi+at to find vi (velocity initial) vf= 0 (because it stops and begins to fall back down)
then use
Use xf=xi + vt + 1/2 at^2 to find xf (the final height) with xi = initial height = 0
vf=0 = vi + (-9.8)( 1.75) yields vi = 17.15 m/s
xf = xi + 17.5(1.75) + 1/2(-9.8)(1.75)^2
xf= 0 + 30.625 - 15.006= 15.619 m
+36915 Tha will always be the case when you neglect air friction and there are no other forces besides gravity acting on the object.
