8. A tennis player hits a ball with a horizontal speed of 30 m s−1. She is standing 4.8 m from the net. How much has the ball dropped vertically by the time it reaches the net?
Solution Time = Distance
___________ Velocity = 4.8
_____ 30 = 0.16 s
Distance fallen in 0.16 s: s = ut + 1 = (0)t + 1
= 0.125 m
9. A bullet is fired vertically upwards at a velocity of 140 m s−1. What is the maximum height the bullet reaches before beginning to fall?
Solution
At the maximum height the bullet’s velocity is zero because it is just about to start falling.
Use v2 = u2 + 2as
(0)2 = (140)2 + 2(−9.8)s (g is negative when acting against velocity) 0 = 19 600 − 19.6s 19.6s = 19 600 s = 1000 m
10. A gun fires a bullet vertically upwards with an initial velocity of 300 m s−1. How many seconds does it take for the bullet to return to the gun height and what distance has it travelled in this time?
Solution
For this question, you need to think of displacement as being the straight line distance from where you started. In this
You may now complete Exercise 4C (page 45).
_ 2 at2
_ 2 (9.8)(0.16)2
case, the bullet has a displacement of 0 if it has returned to where it started (therefore s = 0). You can then find the maximum height and double it for total distance travelled.
s = ut + 1
_ 2 at2 = (0) = (300)(t) + 1
0 = 300t − 4.9t2 0 = t(300 − 4.9t)
Therefore: t = 0 or 300 − 4.9t = 0 t = 0 is the time the bullet left the gun
300 = 4.9t 300
_____ 4.9 = t
61.22s = t (time to return to gun height) Max height means v = 0: v = u + at = 300 + (−9.8)(t)
_______ −9.8 = t
−300 = −9.8t −300
30.61s = t (Note: this is half of the time for the full journey.)
Now substitute for s on the way up: s = ut + 1
_ 2 at2
s = (300)(30.61) + 1
s = 9183.67 − 150 s = 9033.67 m
Double this figure to calculate total distance up and down:
