Chapter 2: Problem 15

A ball is thrown upward with an initial velocity of \(49 \mathrm{~m} / \mathrm{s}\) from \(539 \mathrm{~m}\) high. How high does the ball get, and how long does in take before it hits the ground? [Use results from the simple free fall problem, \(y^{\prime \prime}=-g\).]

Short Answer

Expert verified

The ball will reach a maximum height of 589 meters, and it will take 31.8 seconds for it to hit the ground.

Step by step solution

Initial Setup

The ball is thrown from a height of 539 m with an initial velocity of 49 m/s. Let's let upwards be positive. Therefore, the acceleration due to gravity, \( g \), is -9.8 m/ \( s^2 \). Now we will plug into the position function: \( y(t) = y_0 + v_0*t - 0.5*g*t ^2 = 539 + 49t - 4.9t^2 \).

Calculate Maximum Height

To find the maximum height, we need to find when the ball's velocity is 0 (when it changes direction from moving upwards to falling). This is done by taking the derivative of y(t), which gives \( v(t) = y'(t) = v_0 - g*t = 49 - 9.8t \). Setting this equal to 0 gives \( t = v_0/g = 49/9.8 = 5 \) seconds. To find the maximum height, we substitute \( t = 5 \) back into the position function to get \( y(5) = 539 + 49*5 - 4.9*5^2 = 589 \) meters.

Calculate Total Time of Flight

To find out when the ball hits the ground, we set y(t) = 0. Solving \( 539 + 49t - 4.9t^2 = 0 \) gives two solutions \( t = 31.8 \) and \( t = -21.9 \). Since time cannot be negative, the total time of flight is \( 31.8 \) seconds.