__Question 39__

A ball is thrown horizontally from the top of a building,
as shown in Fig. 2.1.

**Fig. 2.1**

The ball is thrown with a horizontal speed of 8.2 m s^{-1}.
The side of the building is vertical. At point P on the path of the ball, the
ball is distance *x *from the building and is moving at an angle of 60° to
the horizontal. Air resistance is negligible.

**(a) **For the ball at point P,

**(i) **show that the vertical component of its velocity is 14.2
m s^{-1}, [2]

**(ii) **determine the vertical distance
through which the ball has fallen, [2]

**(iii) **determine the horizontal distance *x*.
[2]

**(b) **The path of the ball in **(a)**, with an initial
horizontal speed of 8.2 m s^{-1}, is shown again in Fig. 2.2.

**Fig. 2.2**

On Fig. 2.2, sketch the new path of the ball for the ball
having an initial horizontal speed

**(i) **greater than 8.2 m s^{-1} and with negligible air resistance
(label this path G), [2]

**(ii) **equal to 8.2 m s^{-1} but with air resistance (label this
path A). [2]

**Reference:** *Past Exam Paper – November 2010 Paper 21 Q2*

**Solution:**

**(a) **

**(i) **

{The horizontal component
of the speed is not affected as the air resistance is negligible. The
horizontal component is constant throughout the motion. So, at P, the
horizontal component is still 8.2 m s^{-1}.}

Horizontal speed = 8.2 m s^{-1}

tan 60

^{ o} = v

_{v}
/ 8.2

[tan 60 = opp /
adj]

Vertical component of
speed: v_{v} = 8.2 tan60^{o} = 14.2 m s^{-1}

**(ii) **

{Consider the vertical
motion.

v^{2} = u^{2}
+ 2as = 0 + 2as}

14.2^{2} = 0 + (2×9.8×h)

h = 14.2^{2} / (2×9.8)

Vertical distance fallen:
h = 10.3 m

**(iii) **

{Time is the quantity that
we will use to relate the horizontal and vertical motion.

We first need to determine
the time taken by the ball fall the distance calculated in part (ii) for the
vertical motion.

For the vertical motion,

v = u + at

14.2 = 0 + 9.8t}

Time of descent of ball: t
(= v / a) = 14.2 / 9.8 = 1.45 s

{We can now calculate the
horizontal distance moved by the ball in this amount of time. Since the
horizontal speed is constant, we can use the formula

Speed = distance / time

Distance = Speed × Time}

Horizontal distance: x =
8.2 × 1.45 = 11.9 m

**(b) **

**(i) **

The sketch shows a smooth
path curved and above the given path. The ball hits the ground at a more acute
angle

{The ball would travel a
greater horizontal distance at any instant of time as the horizontal speed is
greater. Note that vertical component of velocity behaves in the same way as
previously. The horizontal component is greater than before (and constant at
all time), so the resultant velocity will be at a more acute angle than
before.}

**(ii) **

The sketch shows a smooth
path curved and below the given path. The ball hits the ground at a steeper
angle.

{Air resistance affects
all components of the speed, reducing them, since it opposes motion. So, a
smaller horizontal distance is travelled by the ball.

The vertical component of
speed undergoes acceleration (due to gravity) until terminal speed is reached (the
ball may hit the ground before terminal velocity is reached) but compare to the
cases before, the speed increases at a smaller rate due to air resistance.

As for the horizontal component, it decreases due to air resistance. So,
the resultant speed will be at a steeper angle to the ground since the
horizontal component is smaller than in the previous cases.}