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Chapter 2: Problem 63

A ball is thrown upward from the ground with an initial speed of \(25 \mathrm{~m} / \mathrm{s}\); at the same instant, another ball is dropped from a building \(15 \mathrm{~m}\) high. After how long will the balls be at the same height?

Short Answer

Expert verified
The two balls will be at the same height after 0.6 seconds.

Step by step solution

01

Establish Initial Conditions

First, establish the initial conditions of each ball. For the ball thrown upwards from the ground, the initial height (\(h_0\)) is 0, the initial speed (\(v_0\)) is +25 m/s, and since the ball is thrown upward, the acceleration (due to gravity, \(g\)) is -9.8 m/s虏. For the ball dropped from the building, the initial height is 15 m, the initial speed is 0 (since it's dropped, not thrown), and the acceleration is downwards, or +9.8 m/s虏.
02

Set up Equations for the Heights of Both Balls

Use the kinematic equation \(h = h_0 + v_0*t - 0.5*g*t^2\) to set up equations for the heights of both balls as functions of time. For ball 1, \(h_1 = 0 + 25t - 0.5*(-9.8)*t^2\), and for ball 2, \(h_2 = 15 + 0*t - 0.5*(9.8)*t^2\).
03

Equate the Two Equations and Solve for 't'

Equate the two equations \(h_1 = h_2\) because we want to find the time at which the balls are at the same height. In other words, set \(0 + 25t - 5*t^2 = 15 - 5*t^2\), which simplifies to \(25t = 15\). Solve for 't' by dividing both sides by 25, which will yield \(t = 15/25 = 0.6 s\).

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Most popular questions from this chapter

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

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