The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
Answer:
Step-by-step explanation:
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<h3>Hope it is helpful....</h3>
Answer:
x+3y-6=0
Step-by-step explanation:
given eqn is y=3x-2 which is 3x-y-2=0
the eqn of line perpendicular to given eqn is -x+3y+k=0
it passes through (6,4)
-6+3*4+k=0
or,. -6+12+k=0
or, k= -6
therefore, the eqn of line perpendicular to given eqn is x+3y-6=0
Answer:
X
−
4.2
Step-by-step explanation:
