Question

1. a golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is the ball’s maximum height? (hint: at the top of its flight, the ball’s vertical velocity component will be zero.) 3. a baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. if the ball is caught 42.0 m from the thrower, how long was it in the air? how high above the thrower did the ball travel?

Explanation:

Problem 2 Solution:

Step 1: Recall projectile motion formulas

The horizontal range \( R \) of a projectile is given by \( R=\frac{v_0^2\sin(2\theta)}{g} \), and the maximum height \( H \) is given by \( H = \frac{v_0^2\sin^2(\theta)}{2g} \), where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g = 9.8\space m/s^2 \) is the acceleration due to gravity. First, we find \( v_0 \) from the range formula.

Given \( R = 301.5\space m \), \( \theta=25.0^\circ \), \( g = 9.8\space m/s^2 \)

From \( R=\frac{v_0^2\sin(2\theta)}{g} \), we can solve for \( v_0^2 \):

\( v_0^2=\frac{Rg}{\sin(2\theta)} \)

\( \sin(2\times25.0^\circ)=\sin(50^\circ)\approx0.7660 \)

\( v_0^2=\frac{301.5\times9.8}{0.7660}=\frac{2954.7}{0.7660}\approx3857.3 \)

Step 2: Calculate maximum height

Now use the maximum height formula \( H=\frac{v_0^2\sin^2(\theta)}{2g} \)

\( \sin(25.0^\circ)\approx0.4226 \), so \( \sin^2(25.0^\circ)\approx0.1786 \)

\( H=\frac{3857.3\times0.1786}{2\times9.8}=\frac{688.9}{19.6}\approx35.15\space m \)

Step 1: Analyze horizontal motion

In projectile motion, the horizontal motion is uniform with velocity \( v_{0x}=v_0\cos\theta \). The horizontal distance \( x = v_{0x}t \), where \( t \) is the time of flight.

Given \( v_0 = 23.0\space m/s \), \( \theta = 25^\circ \), \( x = 42.0\space m \)

\( v_{0x}=23.0\times\cos(25^\circ)\approx23.0\times0.9063 = 20.84\space m/s \)

Step 2: Solve for time \( t \)

From \( x = v_{0x}t \), we get \( t=\frac{x}{v_{0x}} \)

\( t=\frac{42.0}{20.84}\approx2.015\space s \approx 2.02\space s \)

Part 2: Maximum height

Step 1: Analyze vertical motion

The initial vertical velocity \( v_{0y}=v_0\sin\theta \). The maximum height \( H \) is given by \( H=\frac{v_{0y}^2}{2g} \) (since at maximum height, vertical velocity \( v_y = 0 \))

\( v_{0y}=23.0\times\sin(25^\circ)\approx23.0\times0.4226 = 9.7198\space m/s \)

Step 2: Calculate maximum height

\( H=\frac{(9.7198)^2}{2\times9.8}=\frac{94.47}{19.6}\approx4.82\space m \)

Answer:

The maximum height of the golf ball is approximately \( \boldsymbol{35.2\space m} \) (rounded to three significant figures)

Problem 3 Solution:

Part 1: Time in the air