Asked by John
A projectile is fired into the air from the top of a cliff of height h = 208 m above a valley. Its initial velocity is v0 = 55.2 m/s at an angle θ = 54° above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.)
Answers
Answered by
Damon
Do the whole vertical problem first
Vi = 55.2 sin 54
v = Vi - 9.81 t
at top v = 0
so
time to top = (55.2/9.81) sin 54
height at top = 208 + Vi t - 4.9 t^2
do that calculation
now you have t at top and height at top
now it falls from height H at top to ground. How long does that take?
0 = H - 4.9 t^2
t = sqrt (H/4.9)
that is time to fall so
total time in air = time to top + time to fall
NOW you can do the horizontal problem
u = 208 cos 54 the whole time
so
d = 208 cos 54 * total time in air
Vi = 55.2 sin 54
v = Vi - 9.81 t
at top v = 0
so
time to top = (55.2/9.81) sin 54
height at top = 208 + Vi t - 4.9 t^2
do that calculation
now you have t at top and height at top
now it falls from height H at top to ground. How long does that take?
0 = H - 4.9 t^2
t = sqrt (H/4.9)
that is time to fall so
total time in air = time to top + time to fall
NOW you can do the horizontal problem
u = 208 cos 54 the whole time
so
d = 208 cos 54 * total time in air
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