   Chapter 2, Problem 63AP

Chapter
Section
Textbook Problem

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

To determine
The time taken to the ball to reach the same height.

Explanation

Given Info: The initial velocity of the first ball is 25.0m/s and the initial position of the second ball is 15.0m

Explanation:

The formula used to calculate the position of the first ball is,

h=v0t12gt2 (1)

Here,

h is the position at which the balls meet

v0 is the initial velocity of the first ball

a is the acceleration of the first ball

t is the time at which the balls meet

The formula used to calculate the position of the second ball is,

h=h012gt2 (2)

Here,

h0 is the position of the second ball

Compare the equations (1) and (2) for position of the balls.

v0t12gt2=h012gt2

Rearrange the equation to calculate the time at which the balls meet

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