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

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.

v² = u² + 2as = 0 + 2as}

14.2² = 0 + (2×9.8×h)                        

h = 14.2² / (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.}