A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 43.0 m/sat an angle of$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal. The ball strikes the fairway a horizontal distance of180 mfrom the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?

• Initial velocity of the ball, $V_i=43.0\mathrm{m}/\mathrm{s}$.
• Angle at which the ball is above the horizontal,$\theta =30°$.
• The horizontal distance from the tee, $x=180\mathrm{m}$.

The kinematic equations relate the initial velocity, final velocity, displacement, acceleration, and time of motion of the given object.

We can find the height of the fairway and speed at the fairway by using kinematic equations.We can find the components of initial velocity and using that we can find the time from the horizontal distance given. From that time, we can find the rise in height. Using another kinematic equation, we can find the speed at the fairway.

Formulae:

The first kinematic equation of motion, $V_f=V_i+at$ (i)

The second kinematic equation of motion, $d=V_{i}t+\left(\frac{1}{2}\right)a{t}^2$(ii)

Here, $V_i$is the initial velocity, $V_f$is final velocity, $a$is acceleration, $t$is time, and $d$is displacement.

We can resolve velocity in x and y components, we get

$$\begin{gathered} V_{ix}=V_i\cos \theta \\ {V}_{iy}=V_i\sin \theta \end{gathered}$$

The time of flight from a given horizontal distance along the horizontal direction$a_x=0$ . We can substitute the value of displacement in the horizontal direction, initial velocity, and angle in equation (i) to find the time.

$$\begin{gathered} x=V_i\cos \theta \left(t\right) \\ 180\mathrm{m}=43.0\mathrm{m}/\mathrm{s}\left(\cos \left(30°\right)\right)\left(\mathrm{t}\right) \\ \mathrm{t}=\frac{180\mathrm{m}}{43.0\mathrm{m}/\mathrm{s}·\cos \left(30°\right)} \\ =4.83\mathrm{s} \end{gathered}$$

Using this time and equation (ii), the height of rise can be calculated. Substitute the values of initial velocity, time, and acceleration in equation (ii).

$$\begin{gathered} ‐y=V_i\sin \theta \left(t\right)+\left(\frac{1}{2}\right)\left(a\right){\left(t\right)}^2\left(\therefore ∆y=‐y\right) \\ =43.0\mathrm{m}/\mathrm{s}·\sin \left(30°\right)\left(4.83\mathrm{s}\right)+0.5\left(‐9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(4.83\mathrm{s}\right)}^2 \\ y=‐10.46\mathrm{m} \\ \approx ‐11\mathrm{m} \end{gathered}$$

Hence, the value of height rise is 11 m . Negative sign of the rise is because the action is opposite to the acceleration of gravity.

Acceleration along horizontal direction $a_x=0$hence using equation (i), we get the final velocity in x-direction as:

$$\begin{gathered} V_{fx}=43.0\mathrm{m}/\mathrm{s}·\cos \left(30°\right)+0 \\ =37.2390\mathrm{m}/\mathrm{s} \end{gathered}$$

Similarly for final vertical velocity at fairway, we use equation (i). Substitute the values of initial velocity, angle, and gravitational acceleration in equation (i) to calculate the y component of the final velocity.

$$\begin{gathered} V_{fy}=43.0\mathrm{m}/\mathrm{s}·\sin \left(30°\right)+\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(4.83\mathrm{s}\right) \\ =-25.834\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the resultant velocity is calculated by substituting the values of x and y components of the final velocity.

$$\begin{gathered} V_f=\sqrt{\left(V_{fx}^2+V_{fy}^2\right)} \\ =\sqrt{{\left(37.2390\mathrm{m}/\mathrm{s}\right)}^2+{\left(-25.834\mathrm{m}/\mathrm{s}\right)}^2} \\ =45.32\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed at fairwayis 45 m/s .