Question

A golfer practicing on a range with an elevated tee 4.9m above the fairway is able to strike a ball so that it leaves the club with a horizontal velocity of 20ms-1(assume the acceleration due to gravity is 9.80ms-2 and the effects of air resistance may be ignored unless otherwise stated)
(a)how long after the ball leaves the club will it land on the fairway?
(b)what horizontal distance will the ball travel before striking the fairway?
(c)what is the acceleration of the ball 0.5s after being hit?
(d) calculate the speed of the ball 0.8 after it leaves the club
(e) with what speed will the ball strike the ground?

Answer 1

a) The time of falling can be calculated from the kinematic law:

$$h = \dfrac{gt^2}{2}\\ t = \sqrt{\dfrac{2h}{g}}$$

where $h = 4.9m$ is the height and $g = 9.8 m/s^2$. Thus, obtain:

$$t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2\cdot 4.9}{9.8}} = 1 s$$

b) The horisontal distance can be found using the definition of the velocity:

$$v_h = \dfrac{d}{t}$$

where $v_h = 20 m/s$ is the horizontal velocity and $t = 1s$. Since the gravity force acts only in the vertical direction, the horizontal velocity remains constant. Thus:

$$d = vt = 20\cdot 1 = 20m$$

c) The acceleration of the ball at any moment during the falling is constant and equal to 9.80ms-2.

d) The speed of the ball can be found using the Pythagorean theorem:

$$v = \sqrt{v_h^2 + v_v^2}$$

where $v_h = 20m/s$ is constant horizontal velocity (as mentioned in (b)) and $v_v$ is the vertical velocity. It can be found using the definition of the constant acceleration:

$$v_v = gt$$

Thus, since $t = 0.8s$, the vertical velocity will be:

$$v_v = 9.8\cdot0.8 = 7.84 m/s$$

Substituting into the Pythagorean theorem, obtain:

$$v = \sqrt{v_h^2 + v_v^2} = \sqrt{20^2 + 7.84^2} \approx 21.48 m/s$$

e) When the ball strikes the ground (at $t = 1s$ ), it has the following vertical velocity:

$$v_v = gt = 9.8\cdot1 = 9.8m/s$$

Thus, the speed will be:

$$v = \sqrt{20^2 + 9.8^2} \approx 22.27m/s$$

Answer. (a) 1s, (b) 20m, (c) 9.80 ms-2, (d)21.48 m/s, (e)22.27 m/s.