Asked by Anonymous
A boy throws a rock with an initial velocity of 1.67 m/s at 30.0° above the horizontal. If air resistance is negligible, how long does it take for the rock to reach the maximum height of its trajectory?
Vi = 1.67 sin 30
= 0.835 m/s
v = Vi - 4.9 t
v = 0
0 = 0.835 - 4.9 t
solved for t and got 0.17s but apparently its wrong (don't know if i did this right or not) cause the list of answers i could choose from is only
a. 0.119s
b. 0.102s
c. 6.82e-2
d. 8.52e-2
Vi = 1.67 sin 30
= 0.835 m/s
v = Vi - 4.9 t
v = 0
0 = 0.835 - 4.9 t
solved for t and got 0.17s but apparently its wrong (don't know if i did this right or not) cause the list of answers i could choose from is only
a. 0.119s
b. 0.102s
c. 6.82e-2
d. 8.52e-2
Answers
Answered by
Henry
Vo = 1.67m/s[30o].
Yo = 1.67*sin30 = 0.835 m/s.
Y = Yo + g*t = 0.
0.835 - 9.8*t = 0.
9.8t = 0.835.
t = 0.0852 s. = 8.52*10^-2 s.
Yo = 1.67*sin30 = 0.835 m/s.
Y = Yo + g*t = 0.
0.835 - 9.8*t = 0.
9.8t = 0.835.
t = 0.0852 s. = 8.52*10^-2 s.
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