Asked by nabeelah
a gazelle attempts to leap a 1.8 meter fence. assuming a 45 degree takeoff angle, what's the minimum speed for the jump?
Answers
Answered by
Jai
recall that for a projectile motion, the maximum height that the object can reach is given by the formula,
h,max = [(vo)^2 * sin^2 (θ)]/(2g)
where
vo = initial velocity
θ = angle of release
g = acceleration due to gravity = 9.8 m/s^2
the unknown in the problem is vo. thus we just substitute the given in the formula:
h,max = [(vo)^2 * sin^2 (θ)]/(2g)
1.8 = [(vo)^2 * sin^2 (45)]/(2*9.8)
1.8*2*9.8 = (vo)^2 * (1/2)
34.2 = (vo)^2 * (1/2)
70.56 = (vo)^2
vo = sqrt(70.56)
vo = 8.4 m/s
hope this helps~ :)
h,max = [(vo)^2 * sin^2 (θ)]/(2g)
where
vo = initial velocity
θ = angle of release
g = acceleration due to gravity = 9.8 m/s^2
the unknown in the problem is vo. thus we just substitute the given in the formula:
h,max = [(vo)^2 * sin^2 (θ)]/(2g)
1.8 = [(vo)^2 * sin^2 (45)]/(2*9.8)
1.8*2*9.8 = (vo)^2 * (1/2)
34.2 = (vo)^2 * (1/2)
70.56 = (vo)^2
vo = sqrt(70.56)
vo = 8.4 m/s
hope this helps~ :)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.