Asked by Anonymous
a trebuchet launches a projectile with an intial velocity of 15 m/s is launcged at an angle of 35 degrees. what will be the projectiles peak height
Answers
Answered by
Damon
just do vertical problem
initial speed up = Vi = 15 sin 35 (around 8.5 m/s but you do)
v = Vi - g t = Vi - 9.81 t
when at the top,v = 0
so at the top
t = Vi / 9.81 (around .9 but you do)
then
h = Vi t - (1/2) g t^2
h = 8.5 *.9 - 4.9 (.9)^2 (my numbers are approximate)
initial speed up = Vi = 15 sin 35 (around 8.5 m/s but you do)
v = Vi - g t = Vi - 9.81 t
when at the top,v = 0
so at the top
t = Vi / 9.81 (around .9 but you do)
then
h = Vi t - (1/2) g t^2
h = 8.5 *.9 - 4.9 (.9)^2 (my numbers are approximate)
Answered by
henry2,
Vo = 15m/s[35o].
Yo = 15*sin35 = 8.6 m/s. = Vert. component.
Y^2 = Yo^2 + 2g*h.
0 = 8.6^2 + (-19.6)h,
h = 3.77 m.
Yo = 15*sin35 = 8.6 m/s. = Vert. component.
Y^2 = Yo^2 + 2g*h.
0 = 8.6^2 + (-19.6)h,
h = 3.77 m.
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