Question

After a golf ball is hit it takes off with an initial speed of 27.0 m/s and at an angle of 36.0® with respect to the horizontal. The golf field is flat and horizontal. Neglecting air resistance how far will the golf ball fily Stutmt Anawar Tries 0/12 How hiqh will the golf ball rise? Submit Anwer Tries 0/12 How much time will the ball spend in the air? Sutmit Answer Tries 0/12 ow far would the ball fly if the initial speed was doubled? Sutmit Ans Tries 0/12 How much time would the ball spend in the air in this second case? Sutmit Answer Tries Post Discussion

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Initial speed of the ball = V₁ = 27 m/s

Angle of the velocity with the horizontal = $\theta$ = 36^o

Distance the golf ball will fly = R

Range of a projectile is given by,

$R = \frac{V_{1}^{2}Sin2\theta }{g}$

$R = \frac{(27)^{2}Sin(72)}{9.81}$

R = 70.675 m

Maximum height the golf ball reaches = H

$H = \frac{V_{1}^{2}Sin^{2}\theta }{2g}$

$H = \frac{(27)^{2}Sin^{2}(36)}{(2)(9.81)}$

H = 12.84 m

Time the ball will spend in the air = T

$T = \frac{2V_{1}Sin\theta }{g}$

$T = \frac{(2)(27)Sin(36)}{9.81}$

T = 3.235 sec

Now the initial speed is doubled.

New initial speed of the ball = V₂ = 2V₁ = 54 m/s

Distance the ball will fly with the new initial speed = R_n

$R_{n} = \frac{V_{2}^{2}Sin2\theta }{g}$

$R_{n} = \frac{(54)^{2}Sin(72)}{9.81}$

R_n = 282.7 m

Time the ball will spend in air with the new initial speed = T_n

$T_{n} = \frac{2V_{2}Sin\theta }{g}$

$T_{n} = \frac{(2)(54)Sin(36)}{9.81}$

T_n = 6.471 sec

a) Distance the golf ball will fly = 70.675 m

b) Maximum height reached by the golf ball = 12.84 m

c) Time spent in the air by the golf ball = 3.235 sec

d) Distance the golf ball will fly if the initial speed is doubled = 282.7 m

e) Time spent in the air by the golf ball if the initial speed is doubled = 6.471 sec