Question

A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.

Answer 1

Answer:

a. The initial Velocity is 50 m/s approximately.

b. The maximum height at that velocity is 79.8 m.

Explanation:

We are given with:

initial angle = Ф = 34°

Range at that angle = R = 240 m

a. Initial Velocity = v = ?

We know that the range of a projectile is given as:

R = v²sin2Ф/g      where g = 9.8 m/s²

From this equation we can calculate v as well:

v² = R*g/sin2Ф

Putting the values:

v² = 240*9.8/sin (2*34°)

v² = 2352/sin68°

v² = 2352/0.93

v² = 2529

Taking square root:

v = 50 m/s approx

b. Max height reached by the ball if the initial velocity is 50 m/s with initial angle 34° can be calculated as:

H = v²sin²Ф/g

H = 2500*sin²34°/2(9.8)

H = 2500*.3127/19.6

H = 79.77 m

H = 39.9 m

Answer 2

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle$=34^{\circ}$

Range=240 m

We know that Range $=\frac{u^2sin2\theta }{g}$

$240=\frac{u^2sin(68)}{9.81}$

u=50.39 m/s

(b)maximum height of projectile is given by

$H=\frac{u^2sin^2\theta }{2g}$

$H=\frac{50.39^2(sin34)^2}{2\times 9.81}$

H=40.46 m