- #1
faoltaem
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[tex]_{}[/tex]
If a long jumper at the top of his projectory is moving at 6.5 m/s (horizontally) and his cantre of mass is 1.1m above where it was when he launched into the jump, how fast must he have been moving when he launched?
v[tex]_{x}[/tex] = v[tex]_{0x}[/tex] + a[tex]_{x}[/tex]t
x = [tex]\frac{1}{2}[/tex] (v[tex]_{0x}[/tex] + v[tex]_{x}[/tex])t
x = v[tex]_{0x}[/tex]t + [tex]\frac{1}{2}[/tex]a[tex]_{x}[/tex]t[tex]^{2}[/tex]
v[tex]_{x}[/tex][tex]^{2}[/tex] = v[tex]_{0x}[/tex][tex]^{2}[/tex] + 2a[tex]_{x}[/tex]x
v[tex]_{x}[/tex] = 6.5 m/s
y = 1.1m
y[tex]_{0}[/tex] = 0m
while the jumper is at the top of trajectory [tex]\rightarrow[/tex] v[tex]_{y}[/tex] = 0m/s
is it possible to work out this question with just those equations?
they all have either a time, acceleration of x component (i.e. distance travelled)
also is this considered a projectile motion problem as he was running before he jumped.
Homework Statement
If a long jumper at the top of his projectory is moving at 6.5 m/s (horizontally) and his cantre of mass is 1.1m above where it was when he launched into the jump, how fast must he have been moving when he launched?
Homework Equations
v[tex]_{x}[/tex] = v[tex]_{0x}[/tex] + a[tex]_{x}[/tex]t
x = [tex]\frac{1}{2}[/tex] (v[tex]_{0x}[/tex] + v[tex]_{x}[/tex])t
x = v[tex]_{0x}[/tex]t + [tex]\frac{1}{2}[/tex]a[tex]_{x}[/tex]t[tex]^{2}[/tex]
v[tex]_{x}[/tex][tex]^{2}[/tex] = v[tex]_{0x}[/tex][tex]^{2}[/tex] + 2a[tex]_{x}[/tex]x
The Attempt at a Solution
v[tex]_{x}[/tex] = 6.5 m/s
y = 1.1m
y[tex]_{0}[/tex] = 0m
while the jumper is at the top of trajectory [tex]\rightarrow[/tex] v[tex]_{y}[/tex] = 0m/s
is it possible to work out this question with just those equations?
they all have either a time, acceleration of x component (i.e. distance travelled)
also is this considered a projectile motion problem as he was running before he jumped.