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Physics 114 Help!!!??? velocity and distance question (part b)?
Problem #2:
The soccer field is 100m long and 70m wide. Steve Zakuani and Freddie Ljungberg are
both at the center line with Zakuani being at the left sideline and Ljungberg at the right
sideline. Ljungberg wants to pass to Zakuani who is running straight down the sideline
towards the corner. The ball and Zakuani need to arrive at the corner at the same time in
order to make this play successful! Zakuani can run with an average speed of about 9m/s.
Ignore air resistance (drag) and friction for this problem!
a. If Ljungberg wants to kick the ball on the ground, what needs to be the ball’s
velocity?
b. If Ljungberg wants to kick the ball through the air, what is the ball’s initial velocity
and angle with the horizontal?
3 Answers
- PearlsawmeLv 71 decade agoFavorite Answer
a) The diagonal = √ [50^2 + 70^2] = 86.02 m.
If U is the velocity of the ball along the diagonal
U / 86.02 = 9 / 50 and hence U = 15.48 m/s.
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b) If Vis the upward velocity inclined at angle α
V cos α = U = 15.48 m/s.
Range R = [V cos α] * 2 V sin α / g = 86.02 m
V sin α = 86.02 *9.81 / [2*15.48] = 27.26 m/s.
V = √ [15.48 ^2 + 27.26 ^2] = 31.34 m/s
tan α = 27.26 / 15.48
α = 60.4 ˚
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- Doc WLv 61 decade ago
(a)
s = ut gives 50 = 9t so t = 50/9
Diagonal has length sqrt(70^2 + 50^2)
s = ut gives u = s/t = sqrt(70^2 + 50^2)/(50/9) m/s
(b) Vertically
u = VsinA
v = -VsinA
a = -g
t = 50/9 (same as before)
v = u + at gives -VsinA = VsinA - g(50/9)
VsinA = 50g/18.........................(1)
Horizontally
s = D = sqrt((70^2 + 50^2)
u = VcosA
t = 50/9
s = ut gives D = VcosA. (50/9)
VcosA = 9D/50..................(2)
(1)/(2) gives tanA = (50g/18)/(9D/50) = 2 500g/162D
Hence A ~ 60.4 degrees with V ~ 31.34 m/s
Hope they are correct!
The method is fine!
