Asked by zein shion

An arrow is shot at 30.0° angle with the horizontal. It has a velocity of 49 m/s.
a. How high will it go?
b. What is the total flight time (between launch and touching the ground) of the object?
c. What horizontal distance will the arrow travel?

Answers

Answered by Anonymous
Vi = 49 sin 30 = 49/2 initial speed up
v = Vi - g t
v = 0 at top
so
time at top = 49 / 2 g
so total time = 49 / g
u= 49 cos 30
range = u * total time

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