How Long is a Ball Thrown Upwards 3m Above Ground?

[Lavace]

Homework Statement

A ball is thrown vertically upwards with a speed of 10m/s from a point of 2m above the horiztonal ground.
a) Calculate the length of time for which the ball is 3m or more above the ground.

Homework Equations

v^2=u^2 + 2as
v=u+at

The Attempt at a Solution

So there's two ways of doing this, there's on way which my teacher taught us, but I forgot it. Then there's the logical 'sledgehammer' way. Which is:

s=1 v=10 a=-9.8 v=?
Work out the velocity at 1 meter above the ground, which is 3m above the ground then, and then find out the time taken (Using the velocity we found) for the ball to reach its maximum height (v=0) and multiply it by two to find the time for when its falling.
v^2=u^2 + 2as
v= SQRT (u^2 + 2as)
v = SQRT (10^2 + 2x-9.8 x 1) <- Leave in this form to maintain accuracy.
Then using:
(v-u)/a = t (Derived from v=u+at) Where u = SQRT(10^2 + 2x-9.8 x 1) and v=0
[-SQRT(10^2 + 2x-9.8 x 1)] / -9.8 = Time to Reach the top.
Multiply this by two to get the total time which is 1.83s.

Nothing wrong with using this method and it would attain full marks in an exam, but I'd like to learn an easier way of doing it (The way the examiners want us to). Any help would be greatful!.
Thanks in advance.

 

[skeeter]

when the ball is 3 m high, the displacement is 1 m

delta y = (voy)t - 4.9t^2

1 = 10t - 4.9t^2

0 = -1 + 10t - 4.9t^2

solve for the two roots ... the amount of time the ball is 3m or above will be the difference of those two times.

1.935 - 0.105 = 1.83 sec

 

[m2003]

First of all, it's great that you have a logical approach to solving this problem. However, as a scientist, it's important to understand the concepts and equations behind the problem rather than just memorizing a specific method taught by your teacher.

To solve this problem, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the displacement. In this case, u = 10m/s, a = -9.8m/s^2 (since the ball is moving vertically upwards), and s = 3m (since we want to find the time when the ball is 3m or more above the ground).

Plugging these values into the equation, we get:

v^2 = (10m/s)^2 + 2(-9.8m/s^2)(3m)
v^2 = 100m^2/s^2 - 58.8m^2/s^2
v^2 = 41.2m^2/s^2

Taking the square root of both sides, we get:

v = ±6.42m/s

Since we are only interested in the positive value (since the ball is moving upwards), the final velocity is 6.42m/s.

Now, we can use v = u + at to find the time taken for the ball to reach its maximum height (when v = 0). Again, u = 10m/s and a = -9.8m/s^2, so we have:

0 = 10m/s + (-9.8m/s^2)t
-10m/s = -9.8m/s^2t
t = 1.02s

Finally, we can use this time to find the total time for which the ball is 3m or more above the ground. Since the ball is thrown upwards, the time taken to reach the maximum height is the same as the time taken to fall back down to 3m above the ground. Therefore, the total time is 2 x 1.02s = 2.04s.

I hope this explanation helps you understand the concepts and equations behind the problem. Keep up the good work in your studies!