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Chapter 2: Problem 94

A ball is thrown straight upward in the air at a speed of \(15.0 \mathrm{~m} / \mathrm{s} .\) Ignore air resistance. a) What is the maximum height the ball will reach? b) What is the speed of the ball when it reaches \(5.00 \mathrm{~m} ?\) c) How long will it take to reach \(5.00 \mathrm{~m}\) above its initial position on the way up? d) How long will it take to reach \(5.00 \mathrm{~m}\) above its initial position on its way down?

Short Answer

Expert verified
a) The maximum height the ball reaches is 11.25 m. b) The speed of the ball when it's at a height of 5.00 m is 5√5 m/s. c) The time it takes to reach a height of 5.00 m on its way up is (3 + √5)/2 s. d) The time it takes to reach a height of 5.00 m on its way down is (3 - √5)/2 s.

Step by step solution

01

Part (a): Maximum height of the ball

We can use equation 3 for this part, with \(v = 0\), \(u = 15 \mathrm{~m/s}\) and \(a = -10 \mathrm{~m/s^2}\). \(0^2 = 15^2 + 2(-10)s \Rightarrow s = 11.25 \mathrm{~m}\) So, the maximum height the ball reaches is \(11.25 \mathrm{~m}\).
02

Part (b): Speed of the ball at a height of \(5.00 \mathrm{~m}\)

We again use equation 3 with \(s = 5.00 \mathrm{~m}\), \(u = 15 \mathrm{~m/s}\) and \(a = -10 \mathrm{~m/s^2}\). \(v^2 = 15^2 - 2(10)(5) \Rightarrow v^2 = 225 - 100 \Rightarrow v = \sqrt{125} \Rightarrow v = 5\sqrt{5}\mathrm{~m/s}\) So, the speed of the ball when it reaches \(5.00 \mathrm{~m}\) is \(5\sqrt{5}\mathrm{~m/s}\).
03

Part (c): Time to reach \(5.00 \mathrm{~m}\) on the way up

We use equation 2 with \(s = 5.00 \mathrm{~m}\), \(u = 15 \mathrm{~m/s}\) and \(a = -10 \mathrm{~m/s^2}\). \(5 = 15t-\frac{1}{2}(10)t^2 \Rightarrow t^2 -3t + 1 = 0\) We solve the quadratic equation for time: \(t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} \Rightarrow t = \frac{3 \pm \sqrt{5}}{2}\) We are considering the time while the ball is going up, so we choose the positive solution: \(t = \frac{3 + \sqrt{5}}{2}\) So, the time to reach \(5.00 \mathrm{~m}\) on its way up is \(\frac{3 + \sqrt{5}}{2} \mathrm{~s}\).
04

Part (d): Time to reach \(5.00 \mathrm{~m}\) on its way down

As we have found out the time for the upward journey to \(5.00 \mathrm{~m}\), and assuming the motion symmetric, we can simply subtract this time from the total time the ball is in the air. Total time in air, \(T = \frac{2(15)}{10} = 3 \mathrm{~s}\) Time to reach \(5.00 \mathrm{~m}\) on way up, \(t_{up} = \frac{3 + \sqrt{5}}{2} \mathrm{~s}\) Now, time for downward journey to \(5.00 \mathrm{~m}\): \(t_{down} = T - t_{up} = 3 - \frac{3 + \sqrt{5}}{2} = \frac{3 - \sqrt{5}}{2}\mathrm{~s}\) So, the time to reach \(5.00 \mathrm{~m}\) on the way down is \(\frac{3 - \sqrt{5}}{2}\mathrm{~s}\).

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