Vertical Projectile Motion
Δt = 2,04s
vf = - 49,50
Homework
Exercise Q1 to Q2
Homework
Exercise Q3
There are different ways to solve this question
depending where you take your motion from.
OPTION 1
OPTION 2
B
A
C
D
Homework
Exercise Q4 to Q5
Homework
Exercise Q6 to Q7
Homework
Exercise Q8
0,5 m
vf2 = vi2 + 2aΔy
vf2 = (-2)2 + 2(-9,8)(-1,8)
vf = ±6,27 m∙s-1
But vf = -6,27 m∙s-1
vf = vi +aΔt
-6,27 = -2 + (-9,8)Δt
Δt = 0,44 s
vf2 = vi2 + 2aΔy
(0)2 = vi2 + 2(-9,8)(0,9)
vi = ±4,20 m∙s-1
But vi = +4,20 m∙s-1
vi = 4,20 m∙s-1 upwards
FnetΔt = mΔv
Fnet(0,2) = (0,5)[(+4,2 – (-6,27)]
Fnet = 26,18 N
5,88 m.s-1
OPTION 1
vf2 = vi2 + 2aΔy
(-19,6)2 = (5,88)2 + 2(-9,8)Δy
Δy = - 17,84m ∴ Height = 17,84 m
OPTION 2
Area between graph and x-axis for 2,6s
Δy = ½bh + ½bh
½ (0,6)(5,88)+ ½(2,6 – 0,6)(-19,6)
Δy = - 17,84m ∴ Height = 17,84 m
tp = 0,3 + 2,6 = 2,9 s
OPTION 1
Area under graph: Δy = ½bh
=½(0,3)(2,94) = 0,44m
OPTION 2
Δy = viΔt + ½aΔt2
= (2,94)(0,3) + ½(-9,8)(0,3)2
= 0,44 m
6.5
Height of balloon when ball was dropped = 17,84 m
Distance travelled by balloon since ball was dropped:
Δy = vΔt
= (5,88)(2,9)
= 17,05 m
Total height of balloon after 2,9 s
= 17,84 +17,05
= 34,89 m
Maximum height of ball above ground = 0,44m
∴ Distance between balloon and ball
=34,89 – 0,44
= 34,45 m
Homework
Exercise Q9 to Q10