Question

A golfer hits a ball from a tee 4 m above the fairway. The ball leaves the club at an angle of 42° to the horizontal with a speed of 23 m.s-1

A. What maximum height (above the fairway) will the ball reach? Units: m

B. From the moment it is released, how long will the ball take to reach the ground? Units: s

C. Calculate the range of the ball (i.e. the distance R on the figure). Units: m

Answer 1

The ball will follow a projectile motion trajectory. So resolving equations in horizontal (x) and vertical (y) direction:

y VO 0 h -R-

Velocity of ball in x direction = 23 * cos (42^o) = 17.1 m/s

Velocity of ball in y direction = 23 * sin (42^o) = 15.4 m/s

A)

Maximum height above fairway (H) = 4 + max. height achieved by ball in trajectory (s)

For computing s:

Initial velocity along y: uy = 15.4 m/s

Final velocity along y: vy = 0 m/s

Acceleration along y: g = -9.8 m/s²

Using motion equations:

So, the maximum height reached above fairway: H = 4 + 12.1 = 16.1 m

B)

Initial velocity along y: uy = 15.4 m/s

Total displacement along y: sy = 0 m/s (Since, the ball reaches the same y-coordinate from which it starts)

Acceleration along y: g = -9.8 m/s²

Time taken for the complete journey: T = Time taken to return to same y-corodinate from which it start (t) + time taken to hit ground after completing projectile (th)

Using motion equation:

$s_y = u_y*t+\frac{1}{2}*a*t^2$

Substituting values:

$0 = 15.4*t+\frac{1}{2}*-9.8*t^2$

$0 = 15.4+\frac{1}{2}*-9.8*t$

$15.4=4.9*t$

For time taken to hit ground after completing projectile (th):

velocity along y = -15.4 m/s

acceleratioon along y = -9.8 m/s²

Distanec travelled along y = -4 m

Using motion equations:

$s = u*t_h+\frac{1}{2}*a*t_h^2$

$-4 = -15.4*t_h+\frac{1}{2}*-9.8*t_h^2$

Solving the above quadratic equation gives: th = 0.24 sec

Hence, The total time taken by ball to reach the ground after it was released = 3.14 + 0.24 = 3.38 sec

C)

The total range of the ball (R) = Horizontal distance travelled during projectile (H) + Distance travelled to hit ground after completing projectile (Rh)

For computing H:

Time taken to complete projectile = 3.14 sec

Velocity along x-direction = 17.1 m/s

Since, there is no external acceleration in x direction:

H = 3.14 * 17.1 = 53.694 m

For computing Rh:

Velocity along x-direction after completing projectile = 17.1 m/s

Time taken to hit ground after completing projectile = 0.24 sec

Therefore, Rh = 0.24 * 17.1 = 4.104 m

So, The complete range of the ball: R = 53.694 + 4.104 = 57.798 m

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