Question

You throw a ball straight up with an initial velocity of 17.0 m/s. It passes a tree branch on the way up at a height of 9.50 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Give your answer in seconds.

Answer 1

We shall use the second equation of motion for this question. i.e. s = ut + (1/2)at2

For this question, we put s = 9.5 m that is the displacement of the tree branch from ground

9.5 = 17*t + (1/2)*(-9.8)*t2

9.5 = 17*t - 4.9*t2

4.9*t2 - 17*t + 9.5 = 0

Solving this quadratic equation in t using the formula for quadratic equation, we get two values of t, i.e. those values of t at which the ball is at the tree branch. Then the difference of these times is the additional time that is asked in the question.

The two values of t that we get by using formula are t1 = 0.7 s and t2 = 2.77 s

Difference = t1 - t2 = 2.77 - 0.7 = 2.07 s

Therefore, the additional time that will pass before the ball passes the tree branch on the way back down is 2.07s

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