Question

A golf ball with an initial angle of 31° lands exactly 227 m down the range on a level course in a time of 5.285.

(a) Neglecting air friction, what initial speed would achieve this result?
(b) Find the maximum height reached by the ball.

Answer 1

To achieve a horizontal distance of 227m, the golf ball must be launched with an initial speed of 43.03 m/s at an angle of 31°. The maximum height reached by the ball is 136.122 m above the starting point.

Step-by-step explanation:

To find the initial speed of the golf ball, we can use the equations of motion for projectile motion. The horizontal distance traveled by the ball is equal to its initial horizontal velocity multiplied by the time of flight. In this case, the horizontal distance is 227 m and the time of flight is 5.285 s. Therefore, the initial horizontal velocity can be found using the equation:

Horizontal distance = Initial horizontal velocity × time of flight

227 m = Initial horizontal velocity × 5.285 s

Solving for the initial horizontal velocity gives:

Initial horizontal velocity = 227 m / 5.285 s = 43.03 m/s

Since there is no horizontal acceleration, the initial horizontal velocity is equal to the initial speed of the ball.

To find the maximum height reached by the ball, we can use the equation for the vertical component of the motion:

Vertical displacement = Initial vertical velocity × time + 0.5 × acceleration due to gravity × time^2

Since the ball lands at the same height it was initially launched, the vertical displacement is zero. The time of flight is 5.285 s and the acceleration due to gravity is 9.8 m/s^2. Solving for the initial vertical velocity gives:

Initial vertical velocity = -0.5 × acceleration due to gravity × time^2

Initial vertical velocity = -0.5 × 9.8 m/s^2 × (5.285 s)^2

Initial vertical velocity = -0.5 × 9.8 m/s^2 × 27.891225 s^2

Initial vertical velocity = -136.122 m

Therefore, the maximum height reached by the ball is 136.122 m above the starting point.