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Chapter 3: Problem 79

Jason is practicing his tennis stroke by hitting balls against a wall. The ball leaves his racquet at a height of \(60 \mathrm{cm}\) above the ground at an angle of \(80^{\circ}\) with respect to the vertical. (a) The speed of the ball as it leaves the racquet is \(20 \mathrm{m} / \mathrm{s}\) and it must travel a distance of \(10 \mathrm{m}\) before it reaches the wall. How far above the ground does the ball strike the wall? (b) Is the ball on its way up or down when it hits the wall?

Short Answer

Expert verified
Answer: To find the height of the ball above the ground when it strikes the wall, calculate the values from Step 1 to Step 4. The height is the sum of the initial height and the vertical displacement. To determine whether the ball is on its way up or down, calculate the vertical component of the velocity at that time using the formula from Step 5. If the vertical velocity is positive, the ball is going up; if negative, the ball is going down.

Step by step solution

01

Determine the horizontal and vertical components of initial velocity

Since the angle with respect to the vertical is given, we can calculate the angle with respect to the horizontal, which would be \(\theta = 90^{\circ} - 80^{\circ} = 10^{\circ}\). We can now find the horizontal and vertical components of the initial velocity: - Horizontal component (\(v_{0x}\)): \(v_{0x} = v_0 \cdot \cos(\theta) = 20\)m/s \(\cdot \cos(10^{\circ})\) - Vertical component (\(v_{0y}\)): \(v_{0y} = v_0 \cdot \sin(\theta) = 20\)m/s \(\cdot \sin(10^{\circ})\)
02

Calculate the time needed to travel the horizontal distance

Using the horizontal component of the initial velocity, we can calculate the time it takes for the ball to travel \(10 \mathrm{m}\) horizontally. \(t = \frac{d_x}{v_{0x}} = \frac{10 \mathrm{m}}{20 \mathrm{m/s} \cdot \cos(10^{\circ})}\)
03

Find the vertical displacement

To find the vertical displacement (\(d_y\)) at the time when the ball reaches the wall, we use the following kinematic equation: \(d_y = v_{0y}\cdot t - \frac{1}{2}gt^2\) Here, \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)), \(v_{0y}\) is the vertical component of the initial velocity, and \(t\) is the time calculated in step 2. \(d_y = (20\mathrm{m/s}\cdot\sin(10^{\circ}))\cdot\left(\frac{10 \mathrm{m}}{20 \mathrm{m/s} \cdot \cos(10^{\circ})}\right) - \frac{1}{2}(9.81 \mathrm{m/s^2})\left(\frac{10 \mathrm{m}}{20 \mathrm{m/s} \cdot \cos(10^{\circ})}\right)^2\)
04

Calculate the height at which the ball strikes the wall

Now, we need to add the initial height of the ball (\(0.6 \mathrm{m}\)) to the vertical displacement found in Step 3. Height = Initial height + vertical displacement = \(0.6 \mathrm{m} + d_y\)
05

Determine whether the ball is going up or down

To determine whether the ball is going up or down when it hits the wall, we need to check the vertical component of the velocity at the time when the ball hits the wall: \(v_y = v_{0y} - gt\) If \(v_y > 0\), the ball is still going up; if \(v_y < 0\), the ball is on its way down. Substitute the values and solve: \(v_y = 20\mathrm{m/s}\cdot\sin(10^{\circ}) - (9.81 \mathrm{m/s^2})\cdot\left(\frac{10 \mathrm{m}}{20\mathrm{m/s}\cdot\cos(10^{\circ})}\right)\) Now, compute the values from Step 1 to Step 5 to obtain the required information.

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Most popular questions from this chapter

A car travels east at \(96 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h}\). It then travels \(30.0^{\circ}\) east of north at \(128 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h} .\) (a) What is the average speed for the trip? (b) What is the average velocity for the trip?
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