Question

A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.
a. Neglecting air friction, what initial speed would achieve this result?
b. Using the speed determined in item (a), find the maximum height reached by the ball.

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

We have that for the Question it can be said that' the initial speed and the maximum height reached by the ball are

• u=71m/s
• h=78m

From the question we are told

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

a. Neglecting air friction, what initial speed would achieve this result?

b. Using the speed determined in item (a), find the maximum height reached by the ball.

a)

Generally the equation for the initial velocity   is mathematically given as

$u=sqrt{\frac{2g*dx}{sin 68}}\\\\Therefore\\\\\u=sqrt{\frac{2*9.5*240}{sin 68}}$

u=71m/s

b)

Generally the equation for the height   is mathematically given as

$h=\frac{u^2sin^2\theta}{2g}\\\\h=\frac{71^2*sin34^2}{2*9.8}$

h=78m

Therefore

The initial speed and the maximum height reached by the ball are

• u=71m/s
• h=78m

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