Asked by Anonymous
A projectile motion thrown with velocity of 200m/s direction of 53 degree above the horizontal?determine height,range,total time,time at particle reach maximum height?
Answers
Answered by
Damon
solve two problems separately
Problem 1 - vertical
Vi = 200 sin 53
v = Vi - 9.8 t
top when
0 = Vi -9.8 t solve for t
That t is half the time it is in the air
h = Vi t - 4.9 t^2 at time t
total time in air = 2 t
Problem 2, horizontal
u = Vi cos 53 until it crashes
d = u * total time = 2 u t
Problem 1 - vertical
Vi = 200 sin 53
v = Vi - 9.8 t
top when
0 = Vi -9.8 t solve for t
That t is half the time it is in the air
h = Vi t - 4.9 t^2 at time t
total time in air = 2 t
Problem 2, horizontal
u = Vi cos 53 until it crashes
d = u * total time = 2 u t
Answered by
Robert
Max Height = 1,306M
Range = 3,840M
Hangtime/Total Time = 32 seconds
Time to reach max height = 16 seconds
Range = 3,840M
Hangtime/Total Time = 32 seconds
Time to reach max height = 16 seconds
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