Question

golf ball leaves the ground at an angle ∅ and hits a tree while moving horizontally at height h above the ground. If the tree is a horizontal distance b from the point of projection,

(a) Show that 𝑡𝑎𝑛∅=2/ℎ𝑏 .

(b) Given that the initial velocity of the ball is 30m/s and ∅= 30𝑜, find the

(i) Maximum height by the golf ball.

(ii) Rang if its not obstructed by the tree.

Answer 1

(a) We know that initial velocity can be represented by x- and y-components, and the height h and horizontal distance b can be expressed in terms of these components:

$$v_x=v\text{ cos}\theta,\\ v_y=v\text{ sin}\theta.\\\space\\ \text{tan}\theta=\frac{\text{sin}\theta}{\text{cos}\theta}=\frac{v_y}{v_x}.$$

The horizontal component can be expressed in terms of time and range b:

$$v_x=\frac bt.$$

The vertical in terms of height and the same time of flight:

$$v_y=\frac{2h}{t}.$$

Substitute this to the expression for tangent:

$$\text{tan}\theta=\frac{2ht}{tb}=\frac{2h}b.$$

(b i) The maximum height is

$$H=\frac{v_y^2}{2g}=\frac{(v\text{ sin}\theta)^2}{2g}=11.5\text{ m}.$$

(b ii) The total range if it is not obscured by the tree:

$$R=v_xT=v_x·2·\frac{v_y}{g}=2\frac{v^2\text{ cos}\theta\text{sin}\theta}{g}=159\text{ m}.$$