How Do You Calculate the Time a Ball Stays Above a Certain Height?

[lando45]

I have my physics exams on Friday so was just revising some of the stuff I covered earlier in the year. I got this question:

"A ball is thrown vertically upwards with a speed 10ms^-1 from a point 2m above horizontal ground. Calculate the length of time for which the ball is 3m or more above the ground."

I did end up getting the correct answer of 1.83s, but I took an extremely long-winded route. I calculated the total time the ball would be in flight, then subtracted the initial part under 3m and the final part under 3m, which I calculated using the simple equations of motion.

I seem to remember earlier on in the year using a much easier and quicker method to get the answer, but I have forgotten it, and can't seem to find it in my notes. How would you go about solving this question?

 

[teken894]

$$y = y_{0} + v_{y0}t + \frac{1}{2}at^2$$

is the general kinematic equation you want to use,
where in your case:

solve the resulting quadratic equation for t

 

[kyle8921]

Hello,

First of all, congratulations on getting the correct answer for the question! It's great that you are revising and preparing for your exams.

To solve this question, you can use the equation of motion for vertical displacement:

h = h0 + v0t + (1/2)at^2

Where h is the final height, h0 is the initial height, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (which is -9.8 m/s^2 for objects thrown upwards).

In this case, h = 3m, h0 = 2m, v0 = 10m/s, and a = -9.8m/s^2. Plugging these values into the equation, we get:

3 = 2 + 10t + (1/2)(-9.8)t^2

Simplifying this equation, we get:

4.9t^2 + 10t - 1 = 0

Using the quadratic formula, we can solve for t and get two solutions: t = 0.183s and t = -1.83s. Since we are looking for the time when the ball is 3m or more above the ground, we can discard the negative solution and only consider the positive solution, which is 0.183s or approximately 1.83s.

Therefore, the ball is 3m or more above the ground for 1.83 seconds.

I hope this helps and good luck on your exams!