   Chapter 3, Problem 55AP

Chapter
Section
Textbook Problem

A golf ball with an initial speed of 50.0 m/s lands exactly 240 m downrange on a level course, (a) Neglecting air friction, what two projection angles would achieve this result? (b) What is the maximum height reached by the ball, using the two angles determined in part (a)?

(a)

To determine
The two projection angles that would achieve the horizontal distance.

Explanation

The horizontal distance travelled is,

x=v0xt=(v0cosθ)t

Here,

v0 is the initial velocity

θ is the angle of projection

So, the equation for vertical displacement becomes

y=v0yt+12ayt2=(v0sinθ)t12gt2

Here,

v0 is the initial velocity

v0y is the vertical component of the initial velocity

g is the acceleration due to gravity

θ is the angle

When the ball lands at 240m , the y coordinate is zero and the elapsed time is,

0=(v0sinθ)t12gt2t=2v0sinθg

Substituting this time equation in the x=(v0cos

(b)

To determine
The maximum height reached by the ball using the two angles.

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