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Chapter 3: Problem 49

A ball is thrown from a point \(1.0 \mathrm{m}\) above the ground. The initial velocity is \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory.

Short Answer

Expert verified
Answer: The maximum height of the ball above the ground is 5.9 m, and its speed at the highest point of the trajectory is 16.968 m/s.

Step by step solution

01

Resolve the velocity vector into components

The initial velocity of the ball is given as \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. To resolve the velocity vector into its horizontal (\(v_{x}\)) and vertical (\(v_{y}\)) components, we will use the trigonometric expressions for sine and cosine: \(v_x = v\cos(\theta)\) \(v_y = v\sin(\theta)\) \(v_x = 19.6\cos(30^{\circ}) = 16.968 \mathrm{m} / \mathrm{s}\) \(v_y = 19.6\sin(30^{\circ}) = 9.8 \mathrm{m} / \mathrm{s}\)
02

Calculate the time to the highest point

At the highest point of its trajectory, the vertical component of the ball's velocity will be zero. We can use the following equation of motion to find the time it takes for the ball to reach its highest point: \(v_y = u_y - gt\) Rearranging for \(t\): \(t = \frac{u_y - v_y}{g} = \frac{9.8 - 0}{9.81} = 1 \mathrm{s}\)
03

Calculate the maximum height above the ground

Now, we can use another equation of motion to find the highest point (\(h\)) of the ball above the ground: \(h = u_yt - \frac{1}{2}gt^2\) \(h = 9.8(1) - \frac{1}{2}(9.81)(1)^2 = 4.9 \mathrm{m}\) However, the ball is thrown from \(1.0 \mathrm{m}\) above the ground, so the maximum height of the ball above the ground is: \(H = 4.9 + 1 = 5.9 \mathrm{m}\)
04

Calculate the speed at the highest point

Since there is no air resistance, the horizontal component of the velocity remains constant throughout the motion. At the highest point of the trajectory, the vertical component of the velocity is zero, and the speed of the ball is equal to its horizontal component: \(v = v_x = 16.968 \mathrm{m} / \mathrm{s}\) Now, we have the answers to both parts of the problem: (a) The maximum height of the ball above the ground is \(5.9 \mathrm{m}\). (b) The speed of the ball at the highest point of the trajectory is \(16.968 \mathrm{m} / \mathrm{s}\).

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