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Chapter 4: Problem 74

With what initial velocity must an object be thrown upward (from a height of 2 meters) to reach a maximum height of 200 meters?

Short Answer

Expert verified
As a result, the object must be thrown upward with an initial velocity of approximately 62.3 m/s to reach a maximum height of 200 meters.

Step by step solution

01

Identifying Given Values

The starting point is to identify the given values from the problem. Here, the initial height (\(d_i\)) is 2 m and the final height (\(d_f\)) is 200 m. Gravity (\(g\)) acts as the acceleration and its value is -9.8 m/s虏, direction is negative as it acts in downward direction, opposite to the motion of the projectile. The final velocity (\(v_f\)) at the maximum height of the projectile is 0 m/s because the object momentarily stops at the peak of its trajectory before descending.
02

Applying the Kinematic Equation

To calculate the initial velocity (\(v_i\)), we use the kinematic equation that connects final velocity, initial velocity, acceleration, and distance: \(v_f^2 = v_i^2 + 2*a*d\). In this case, \(v_f = 0 m/s\), \(a = -9.8 m/s^2\), and \(d = d_f - d_i = 200m - 2m = 198m\). Since we want to find \(v_i\), we rearrange the equation to solve for it: \(v_i = sqrt(v_f^2 - 2*a*d)\).
03

Calculation

Substitute the known values into equation,\(v_i = sqrt(0 - 2*(-9.8)*198) = sqrt(3880.8) m/s\).

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