Asked by Anonymous
A stuntman drives a motorcycle off a 350 m cliff going 70 mph. The angle of elevation of the cliff is 21 degree. He is hoping to make it across a 261 m wide river and land on a ledge 82 m high. Does he make it?
Answers
Answered by
Henry
Xo = 70mi/h * 1600m/mi + 1h/3600s = 31.1 m/s.
h = 350 - 82 = 268 m Above the ledge.
h = 0.5g*t^2 = 268 m.
4.9t^2 = 268
t^2 = 54.69
Tf = 7.4 s. = Fall time.
Dx = Xo*Tf = 31.1m/s * 7.4s = 230 m. =
Hor. distance.
h = 350 - 82 = 268 m Above the ledge.
h = 0.5g*t^2 = 268 m.
4.9t^2 = 268
t^2 = 54.69
Tf = 7.4 s. = Fall time.
Dx = Xo*Tf = 31.1m/s * 7.4s = 230 m. =
Hor. distance.
Answered by
Question!
Thanks for your answer Henry but where did you get the 1600m/mi?
Answered by
Anon
To answer your question 5 years later, thats probably just a mistake on his part lmao
Answered by
M0609
To answer the question from six years ago, the 1600 m is actually supposed to be 1609.344 , because 1 mile = 1609.344 meters. In other problems some people prefer to use 1609 m instead of 1609.344
So, the answer should actually be Xo= 31.2861 m/s since 70mi/h * 1609m/mi + h/3600s= 31.2861m/s
So, the answer should actually be Xo= 31.2861 m/s since 70mi/h * 1609m/mi + h/3600s= 31.2861m/s
Answered by
Anonymous
Does he make it or not?
Answer
how did you get the fall time?
Answered by
Bot
To find the fall time, we use the formula:
h = 0.5gt^2
where h = vertical displacement (difference in height), g = acceleration due to gravity (9.8 m/s^2), and t = time of fall.
Plugging in the values we have:
268m = 0.5(9.8 m/s^2)t^2
Simplifying and solving for t:
t^2 = 54.6939
t ≈ 7.4 seconds
So the fall time is approximately 7.4 seconds.
h = 0.5gt^2
where h = vertical displacement (difference in height), g = acceleration due to gravity (9.8 m/s^2), and t = time of fall.
Plugging in the values we have:
268m = 0.5(9.8 m/s^2)t^2
Simplifying and solving for t:
t^2 = 54.6939
t ≈ 7.4 seconds
So the fall time is approximately 7.4 seconds.
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